LeetCode每日一题

LeetCode每日一题

✨2024-12-27

📙1. 两数之和

法一：遍历数组

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public int[] twoSum(int[] nums, int target) {
    for (int i = 0; i < nums.length - 1; i++) {
        for (int j = i + 1; j < nums.length; j++) {
            if (nums[i] + nums[j] == target) {
                return new int[]{i, j};
            }
        }
    }
    return null;
}

法二：利用哈希表

num为key

index为value

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public int[] twoSum(int[] nums, int target) {
    HashMap<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        if (map.containsKey(target - nums[i])) {
            return new int[]{map.get(target - nums[i]), i};
        }
        map.put(nums[i], i);
    }
    return null;
}

📙175. 组合两个表

法一：左联表

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# Write your MySQL query statement below
select FirstName,LastName,City,State
from Person left join Address 
on Person.PersonId = Address.PersonId;

✨2024-12-28

📙2. 两数相加

法一：合并两数且遍历

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public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode head = new ListNode();
    ListNode tail = head;
    while (l1 != null && l2 != null) {
        ListNode node = new ListNode(l1.val + l2.val);
        tail.next = node;
        tail = tail.next;
        l1 = l1.next;
        l2 = l2.next;
    }
    if (l1 != null) {
        tail.next = l1;
    }
    if (l2 != null) {
        tail.next = l2;
    }
    tail = head.next;
    int add = 0,sum = 0;
    ListNode tailtmp = head;
    while (tail != null) {
        sum = tail.val + add;
        tail.val = sum % 10;
        add = sum / 10;
        tailtmp = tail;
        tail = tail.next;
    }

    if (add > 0) {
        tailtmp.next = new ListNode(add);
    }
    head = head.next;
    return head;
}

📙176. 第二高薪水

法一：排除最高薪水再次搜索最大薪水

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# Write your MySQL query statement below
select MAX(salary) as `SecondHighestSalary`
from Employee
where salary<(select MAX(salary)
                from Employee);

法二：排序后取第二高薪水

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# Write your MySQL query statement below
select (select distinct salary
        from Employee
        order by salary desc
        limit 1 offset 1
        ) as `SecondHighestSalary`;

✨2024-12-29

📙3. 无重复字符的最长字串

法一：通过HashSet判断

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public int lengthOfLongestSubstring(String s) {
    if (s == null || s.length() == 0) return 0;
    HashSet<Character> chars = new HashSet<Character>();
    int max = 1, tmp = 0;
    for (int l = 0; l < s.length() - 1; l++) {
        for (int r = l; r < s.length(); r++) {
            if (chars.contains(s.charAt(r))) {
                tmp = 0;
                chars.clear();
                break;
            } else {
                chars.add(s.charAt(r));
                tmp++;
                if (tmp > max) {
                    max = tmp;
                }
            }
        }
    }
    return max;
}

📙177. 第N高薪水

法一：排序后获取第N高薪水

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CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
    set N := N-1;
  RETURN (
      # Write your MySQL query statement below.
    select distinct salary
    from Employee
    order by salary desc
    limit 1 offset N
  );
END

✨2024-12-30

📙4. 寻找两个正序数组的中位数

法一：合并并排序后找中位数

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public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    int[] nums = new int[nums1.length + nums2.length];
    for (int i = 0; i < nums1.length; i++) {
        nums[i] = nums1[i];
    }
    for (int i = 0; i < nums2.length; i++) {
        nums[nums1.length + i] = nums2[i];
    }
    Arrays.sort(nums);
    int len = nums.length;
    if (len % 2 == 0) {
        return (nums[len / 2 - 1] + nums[len / 2]) / 2.0;
    } else {
        return nums[len / 2];
    }
}

📙178. 分数排名

法一：使用dense_rank()排序

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# Write your MySQL query statement below
select score, dense_rank() over (order by score desc) as 'rank'
from Scores;

收获：

函数 相同值处理 排名跳跃
ROW_NUMBER() 每行唯一编号 无
RANK() 相同值相同排名，后续跳过 有
DENSE_RANK() 相同值相同排名，后续不跳过 无
NTILE(n) 将数据划分为 n 组 无

✨2024-12-31

📙5. 最长回文子串

法一：从中间向两边遍历

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public String longestPalindrome(String s) {
    int max = 0,start = 0;
    int l,r;
    for (int i = 0; i < s.length(); i++) {

        //aba 单数
        l = i;
        r = i;
        while (l >= 0 && r < s.length() && s.charAt(l) == s.charAt(r)) {
            if (r - l + 1 > max) {
                max = r - l + 1;
                start = l;
            }
            l--;
            r++;
        }

        //abba 双数
        l = i;
        r = i + 1;
        while (l >= 0 && r < s.length() && s.charAt(l) == s.charAt(r)) {
            if (r - l + 1 > max) {
                max = r - l + 1;
                start = l;
            }
            l--;
            r++;
        }
    }
    return s.substring(start, start + max);
}

📙180. 连续出现的数字

法一：内连接

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# Write your MySQL query statement below
select distinct l1.num as ConsecutiveNums
from Logs l1,Logs l2,Logs l3
where l1.id-1 = l2.id
and l2.id-1 = l3.id
and l1.num = l2.num
and l2.num = l3.num;

✨2025-1-1

📙6. Z字形变换

法一：根据字形形成的数据创建集合

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public String convert(String s, int numRows) {
    if (numRows == 1) return s;

    int divisor = (numRows - 1) * 2;
    ArrayList<StringBuilder> list = new ArrayList<>();
    for (int i = 0; i < numRows; i++) {
        list.add(new StringBuilder());
    }
    for (int i = 0; i < s.length(); i++) {
        if (i % divisor < numRows) {
            list.get(i % divisor).append(s.charAt(i));
        } else {
            list.get((numRows - 1) - (i % divisor - (numRows - 1))).append(s.charAt(i));
        }
    }
    StringBuilder string = new StringBuilder();
    for (int i = 0; i < numRows; i++) {
        string.append(list.get(i));
    }
    return string.toString();
}

📙181. 超过经理收入的员工

法一：连表查询

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# Write your MySQL query statement below
select e1.name as 'Employee'
from Employee e1
join Employee e2 on e1.managerId=e2.id
where e2.salary < e1.salary;

法二：子查询

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# Write your MySQL query statement below
select name as 'Employee'
from Employee e1
where managerId is not null  
and salary > (select salary
                from Employee e2
                where e1.managerId = e2.id);

✨2025-1-2

📙7. 整数反转

法一：移动尾数

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public int reverse(int x) {
    int sum = 0, tmp = 0;
    while (x != 0) {
        tmp = x % 10;
        if (sum > 214748364 || (sum == 214748364 && tmp > 7)) {
            return 0;
        }
        if (sum < -214748364 || (sum == -214748364 && tmp < -8)) {
            return 0;
        }
        sum = sum * 10 + tmp;
        x /= 10;
    }
    return sum;
}

📙182. 查找重复的电子邮箱

法一：使用Having，group by

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# Write your MySQL query statement below
select email as Email
from Person
group by email
having COUNT(id) >= 2;

✨2025-1-3

📙8. 字符串转换整数 (atoi)

法一：去空格后遍历，之后判断是否超出范围

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public int myAtoi(String s) {
    if (s == null || s.equals("")) {
        return 0;
    }
    s = s.trim();
    boolean negative = false;
    if (s.length() >= 1 && (s.charAt(0) == '-' || s.charAt(0) == '+')) {
        negative = s.charAt(0) == '-';
        s = s.substring(1, s.length());
    }
    if (s.length() < 1 || '0' > s.charAt(0) || s.charAt(0) > '9') {
        return 0;
    }
    int i = 0,result = 0;
    while (i < s.length() && '0' <= s.charAt(i) && s.charAt(i) <= '9') {
        int tmp = s.charAt(i) - '0';
        if (negative) tmp *= -1;
        if (result > Integer.MAX_VALUE / 10 || (result == Integer.MAX_VALUE / 10 && tmp > 7)) {
            return Integer.MAX_VALUE;
        }
        if (result < Integer.MIN_VALUE / 10 || (result == Integer.MIN_VALUE / 10 && tmp < -8)) {
            return Integer.MIN_VALUE;
        }
        result = result * 10 + tmp;
        i++;
    }
    return result;
}

📙183. 从不订购的用户

法一：子查询

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# Write your MySQL query statement below
select name as 'Customers'
from Customers
where id not in (select customerId
                 from Orders);

法二：连表查询

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# Write your MySQL query statement below
select name as 'Customers'
from Customers
left join Orders on Customers.id = Orders.customerId
where Orders.customerId is null

✨2025-1-4

📙9. 回文数

法一：转换为字符数组，左右对比

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public boolean isPalindrome(int x) {
    if (x < 0) return false;
    char[] chars = String.valueOf(x).toCharArray();
    int left = 0, right = chars.length - 1;
    while (left < right) {
        if (chars[left] != chars[right]) {
            return false;
        }
        left++;
        right--;
    }
    return true;
}

法二：获取原数的回文数，判断两数是否相同

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public boolean isPalindrome(int x) {
    if (x < 0) return false;
    int sum = 0, tmp = x;
    while (tmp != 0) {
        sum = sum * 10 + tmp % 10;
        tmp = tmp / 10;
    }
    return sum == x;
}

📙184. 部门工资最高的员工

法一：连表查询

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# Write your MySQL query statement below
select Department.name as 'Department',sub_quary.name as 'Employee',salary as 'Salary'
from (select *,dense_rank() over (partition by departmentId order by salary desc) as salaryId
        from Employee) as sub_quary
left join Department on Department.id = sub_quary.departmentId
where salaryId = 1;
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# Write your MySQL query statement below
select Department.name 'Department',Employee.name 'Employee',salary 'Salary'
from Employee,Department
where Employee.departmentId = Department.id
and (Employee.departmentId, salary) in (select departmentId,MAX(salary)
                                        from Employee
                                        group by departmentId);

✨2025-1-5

📙13. 罗马数字转整数

法一：前数大于后数则加，反之则减

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public int romanToInt(String s) {
    int pre = getInt(s.charAt(0)), num, sum = 0;
    for (int i = 1; i < s.length(); i++) {
        num = getInt(s.charAt(i));
        if (pre < num) {
            sum -= pre;
        } else {
            sum += pre;
        }
        pre = num;
    }
    sum += pre;
    return sum;
}

private int getInt(char c) {
    switch (c) {
        case 'I':
            return 1;
        case 'V':
            return 5;
        case 'X':
            return 10;
        case 'L':
            return 50;
        case 'C':
            return 100;
        case 'D':
            return 500;
        case 'M':
            return 1000;
        default:
            return 0;
    }
}

📙185. 部门工资前三高的所有员工

法一：连表查询

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# Write your MySQL query statement below
select Department.name as 'Department',sub_quary.name as 'Employee',salary as 'Salary'
from (select *,dense_rank() over (partition by departmentId order by salary desc) as salaryId
        from Employee) as sub_quary
left join Department on Department.id = sub_quary.departmentId
where salaryId <= 3;

✨2025-1-6

📙12. 整数转罗马数字

法一：罗列数字，从大到小转换

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public String intToRoman(int num) {
    StringBuilder s = new StringBuilder();
    while (true) {
        if (num >= 1000) {
            s.append("M");
            num -= 1000;
        } else if (num >= 900) {
            s.append("CM");
            num -= 900;
        } else if (num >= 500) {
            s.append("D");
            num -= 500;
        } else if (num >= 400) {
            s.append("CD");
            num -= 400;
        } else if (num >= 100) {
            s.append("C");
            num -= 100;
        } else if (num >= 90) {
            s.append("XC");
            num -= 90;
        } else if (num >= 50) {
            s.append("L");
            num -= 50;
        } else if (num >= 40) {
            s.append("XL");
            num -= 40;
        } else if (num >= 10) {
            s.append("X");
            num -= 10;
        } else if (num >= 9) {
            s.append("IX");
            num -= 9;
        } else if (num >= 5) {
            s.append("V");
            num -= 5;
        } else if (num >= 4) {
            s.append("IV");
            num -= 4;
        } else if (num >= 1) {
            s.append("I");
            num -= 1;
        } else {
            break;
        }
    }
    return s.toString();
}

📙196. 删除重复的电子邮箱

法一：子查询

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# Write your MySQL query statement below
delete 
from Person
where id in (select id
             from (select id,row_number() over (partition by email order by id) as mrank
                    from Person) as sub_quary
             where mrank != 1);

✨2025-1-7

📙19. 删除链表的倒数第 N 个结点

法一：遍历找到结点总数，再找倒数第n个节点

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public ListNode removeNthFromEnd(ListNode head, int n) {
    ListNode tmp = head;
    int len = 0;
    while (tmp != null) {
        tmp = tmp.next;
        len++;
    }
    tmp = head;
    if (len - n - 1 < 0) {
        head = head.next;
        tmp = head;
    } else {
        for (int i = 0; i < len - n - 1; i++) {
            tmp = tmp.next;
        }
        tmp.next = tmp.next.next;
    }
    return head;
}

📙197. 上升的温度

法一：连表查询

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# Write your MySQL query statement below
select d1.id
from Weather d1,Weather d2
where datediff(d1.recordDate, d2.recordDate) = 1
and d1.temperature > d2.temperature;

✨2025-1-8

📙20. 有效的括号

法一：使用栈

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public boolean isValid(String s) {
    char[] chars = s.toCharArray();
    Stack<Character> stack = new Stack<>();
    for (char c : chars) {
        if (c == '(' || c == '{' || c == '[') {
            stack.push(c);
        }
        else if (c == ')' || c == '}' || c == ']') {
            if (stack.isEmpty()) {
                return false;
            }
            if (c == ')') {
                if (stack.peek() == '(') {
                    stack.pop();
                } else {
                    return false;
                }
            } else if (c == '}') {
                if (stack.peek() == '{') {
                    stack.pop();
                } else {
                    return false;
                }
            } else if (c == ']') {
                if (stack.peek() == '[') {
                    stack.pop();
                } else {
                    return false;
                }
            }
        }
    }
    return stack.isEmpty();
}

📙511. 游戏玩法分析 I

法一：子查询

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# Write your MySQL query statement below
select player_id,event_date as first_login
from (select player_id,event_date,row_number() over (partition by player_id order by event_date) as p_rank
    from Activity) as sub_quary
where p_rank = 1;

✨2025-1-9

📙21. 合并两个有序链表

法一：遍历两个链表，依次加入

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public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
    if (list1 == null) return list2;
    if (list2 == null) return list1;

    ListNode head, tmp;
    if (list1.val > list2.val) {
        head = list2;
        tmp = list2;
        list2 = list2.next;
    } else {
        head = list1;
        tmp = list1;
        list1 = list1.next;
    }

    while (list1 != null && list2 != null) {
        if (list1.val > list2.val) {
            tmp.next = list2;
            tmp = tmp.next;
            list2 = list2.next;
        } else {
            tmp.next = list1;
            tmp = tmp.next;
            list1 = list1.next;
        }
    }
    if (list1 != null) {
        tmp.next = list1;
    }
    if (list2 != null) {
        tmp.next = list2;
    }
    return head;
}

📙577. 员工奖金

法一：连表查询

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# Write your MySQL query statement below
select name,bonus
from Employee
left join Bonus on Employee.empId = Bonus.empId
where bonus < 1000
or bonus is null;

✨2025-1-10

📙26. 删除有序数组中的重复项

法一：遍历，通过计数改变原有项

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public int removeDuplicates(int[] nums) {
    int count = 1, tmp = nums[0];
    for (int i = 0; i < nums.length; i++) {
        if (nums[i] != tmp) {
            tmp = nums[i];
            nums[count++] = nums[i];
        }
    }
    return count;
}

📙584. 寻找用户推荐人

法一：is null

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# Write your MySQL query statement below
select name
from Customer
where referee_id is null
or referee_id != 2;

✨2025-1-11

📙27. 移除元素

法一：遍历删除

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public int removeElement(int[] nums, int val) {
    int count = 0, len = nums.length;
    for (int i = 0; i < len; i++) {
        if (nums[i] == val) {
            for (int j = i; j < nums.length - 1; j++) {
                nums[j] = nums[j + 1];
            }
            len--;
            i--;
        } else {
            count++;
        }
    }
    return count;
}

📙570. 至少有5名直接下属的经理

法一：连表查询

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# Write your MySQL query statement below
select e1.name
from Employee e1
left join Employee e2 on e1.id = e2.managerId
group by e1.id
having COUNT(e1.id) >= 5;

✨2025-1-12

📙28. 找出字符串中第一个匹配项的下标

法一：遍历

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public int strStr(String haystack, String needle) {
    if (haystack == null || needle == null || haystack.length() < needle.length()) {
        return -1;
    }

    boolean found;
    for (int i = 0; i < haystack.length() - needle.length() + 1; i++) {
        found = false;
        for (int j = 0, hindex = i; j < needle.length() && hindex < haystack.length(); j++, hindex++) {
            if (haystack.charAt(hindex) == needle.charAt(j)) {
                found = true;
            } else {
                found = false;
                break;
            }
        }
        if (found) {
            return i;
        }
    }
    return -1;
}

优化：

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public int strStr(String haystack, String needle) {
    int hlen = haystack.length(), nlen = needle.length();
    char[] h = haystack.toCharArray(), n = needle.toCharArray();
    for (int i = 0; i <= hlen - nlen; i++) {
        int hindex = i, nindex = 0;
        while (nindex < nlen && h[hindex] == n[nindex]) {
            hindex++;
            nindex++;
        }
        if (nindex == nlen) {
            return i;
        }
    }
    return -1;
}

📙586. 订单最多的客户

法一：子查询

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# Write your MySQL query statement below
select customer_number
from (select customer_number,COUNT(order_number) as nums
    from Orders
    group by customer_number
    order by nums desc) as sub_quary
limit 1;

法二：普通查询

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# Write your MySQL query statement below
select customer_number
from Orders
group by customer_number
order by COUNT(order_number) desc
limit 1;

✨2025-1-13

📙35. 搜索插入位置

法一：二分查找

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public int searchInsert(int[] nums, int target) {
    return search(nums, target, 0, nums.length - 1);
}

public int search(int[] nums, int target, int start, int end) {
    int mid = (start + end) / 2;
    if (nums[mid] == target) {
        return mid;
    }
    if (start == end) {
        if (start == 0) {
            if (nums[start] < target){
                return 1;
            } else {
                return 0;
            }
        } else if (start == nums.length - 1) {
            if (nums[start] < target) {
                return start + 1;
            } else {
                return start;
            }
        } else {
            return mid;
        }
    }
    if (nums[mid] < target) {
        return search(nums, target, mid + 1, end);
    }
    return search(nums, target, start, mid);
}

📙595. 大的国家

法一：普通查找

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# Write your MySQL query statement below
select World.name,World.population,World.area
from World
where World.area >= 3000000
or World.population >= 25000000;

✨2025-1-14

📙58. 最后一个单词的长度

法一：遍历计数

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public int lengthOfLastWord(String s) {
    int len = 0, max = 0;
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == ' ') {
            if (len > 0) {
                max = len;
            }
            len = 0;
        } else {
            len++;
        }
    }
    if (len > 0) {
        max = len;
    }
    return max;
}

📙596. 超过5名学生的课

法一：普通查找

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# Write your MySQL query statement below
select class
from Courses
group by class
having COUNT(student) >= 5;

✨2025-1-15

📙53. 最大子数组和

法一：动态规划

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public int maxSubArray(int[] nums) {
    int preSum = 0, maxSum = nums[0];
    for (int num : nums) {
        preSum = Math.max(preSum + num, num);
        maxSum = Math.max(maxSum, preSum);
    }
    return maxSum;
}

📙607. 销售员

法一：嵌套查询

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# Write your MySQL query statement below
select name
from SalesPerson
where sales_id not in (select sales_id
                    from Orders
                    where com_id in (select com_id
                                    from Company
                                    where name = 'RED'));

✨2025-1-16

📙54. 螺旋矩阵

法一：设置4个标识后按顺序遍历

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public List<Integer> spiralOrder(int[][] matrix) {
    ArrayList<Integer> arrayList = new ArrayList<>();
    int l = 0, r = matrix[0].length - 1;
    int h = 0, t = matrix.length - 1;
    while ((l <= r) && (h <= t)) {
        if (h <= t) {
            for (int i = l; i <= r; i++) {
                arrayList.add(matrix[h][i]);
            }
            h++;
        }
        if (l <= r) {
            for (int i = h; i <= t; i++) {
                arrayList.add(matrix[i][r]);
            }
            r--;
        }
        if (h <= t) {
           for (int i = r; i >= l; i--) {
                arrayList.add(matrix[t][i]);
            }
            t--;
        }
        if (l <= r) {
            for (int i = t; i >= h; i--) {
                arrayList.add(matrix[i][l]);
            }
            l++;
        }
    }
    return arrayList;
}

📙610. 判断三角形

法一：使用if判断

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# Write your MySQL query statement below
select *,if(((x + y) > z and (x + z) > y and (y + z) > x),'Yes','No') as triangle
from Triangle;

✨2025-1-17

📙59. 螺旋矩阵 II

法一：同螺旋矩阵，设置标识后遍历

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public int[][] generateMatrix(int n) {
    int num = 1;
    int[][] matrix = new int[n][n];
    int l = 0, r = n - 1;
    int h = 0, t = n - 1;
    while ((l <= r) && (h <= t)) {
        for (int i = l; i <= r; i++) {
            matrix[h][i] = num++;
        }
        h++;
        for (int i = h; i <= t; i++) {
            matrix[i][r] = num++;
        }
        r--;
        for (int i = r; i >= l; i--) {
            matrix[t][i] = num++;
        }
        t--;
        for (int i = t; i >= h; i--) {
            matrix[i][l] = num++;
        }
        l++;
    }
    return matrix;
}

📙619. 只出现一次的最大数字

法一：子查询

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# Write your MySQL query statement below
select (select num
        from MyNumbers
        group by num
        having COUNT(*) = 1
        order by num desc
        limit 1
) num;

✨2025-1-18

📙61. 旋转链表

法一：将单向连表变为螺旋链表再切断

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public ListNode rotateRight(ListNode head, int k) {
    if (head == null || k == 0) {
        return head;
    }

    int len = 0;
    ListNode tmp = head;
    while (tmp != null) {
        tmp = tmp.next;
        len++;
    }

    tmp = head;
    while (tmp.next != null) {
        tmp = tmp.next;
    }
    tmp.next = head;

    k = k % len;
    tmp = head;
    for (int i = 0; i < len - k - 1; i++) {
        tmp = tmp.next;
    }
    head = tmp.next;
    tmp.next = null;

    return head;
}

📙620. 有趣的电影

法一：普通查找

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# Write your MySQL query statement below
select *
from cinema
where description != 'boring'
and id % 2 != 0
order by rating desc;

✨2025-1-19

📙66. 加一

法一：加一后遍历，如果最高位加一则新建数组

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public int[] plusOne(int[] digits) {
    digits[digits.length - 1]++;
    for (int i = digits.length - 1; i > 0; i--) {
        if (digits[i] >= 10) {
            digits[i] %= 10;
            digits[i - 1] += 1;
        }
    }
    if (digits[0] >= 10) {
        digits[0] %= 10;
        int[] ints = new int[digits.length + 1];
        ints[0] = 1;
        for (int i = 0; i < digits.length; i++) {
            ints[i + 1] = digits[i];
        }
        return ints;
    }
    return digits;
}

📙627. 变更性别

法一：使用条件判断语句

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# Write your MySQL query statement below
update Salary
set sex = (case sex
            when 'm' then 'f'
            else 'm'
            end);

✨2025-1-20

📙67. 二进制求和

法一：首部不够则添0，遍历

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public String addBinary(String a, String b) {
    StringBuilder str = new StringBuilder();
    int carry = 0;
    for (int i = a.length() - 1, j = b.length() - 1; i >= 0 || j >= 0; i--, j--) {
        int sum = carry;
        sum += i >= 0 ? a.charAt(i) - '0' : 0;
        sum += j >= 0 ? b.charAt(j) - '0' : 0;
        str.append(sum % 2);
        carry = sum / 2;
    }
    str.append(carry == 1 ? carry : "");
    return str.reverse().toString();
}

📙1050. 合作过至少三次的演员和导演

法一：通过 group 普通查找

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# Write your MySQL query statement below
select actor_id,director_id
from ActorDirector
group by actor_id,director_id
having COUNT(*) >= 3;

✨2025-1-21

📙69. x 的平方根

法一：袖珍计算器算法

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public int mySqrt(int x) {
    if (x == 0 || x == 1) {
        return x;
    }

    int result = (int) Math.exp(0.5 * Math.log(x));
    return (long) (result + 1) * (result + 1) <= x ? result + 1 : result;
}

📙1068. 产品销售分析 I

法一：连表查询

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# Write your MySQL query statement below
select product_name,Sales.year,price
from Sales
left join Product on Sales.product_id = Product.product_id;

✨2025-1-22

📙73. 矩阵置零

法一：遍历存位置

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public void setZeroes(int[][] matrix) {
    ArrayList<ArrayList<Integer>> list = new ArrayList<>();
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix[i].length; j++) {
            if (matrix[i][j] == 0) {
                ArrayList<Integer> subList = new ArrayList<>();
                subList.add(i);
                subList.add(j);
                list.add(subList);
            }
        }
    }
    for (int z = 0; z < list.size(); z++) {
        ArrayList<Integer> integers = list.get(z);
        int row = integers.get(0);
        int col = integers.get(1);
        for (int i = 0; i < matrix[0].length; i++) {
            matrix[row][i] = 0;
        }
        for (int i = 0; i < matrix.length; i++) {
            matrix[i][col] = 0;
        }
    }
}

优化空间复杂度：

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public void setZeroes(int[][] matrix) {
    int m = matrix.length, n = matrix[0].length;
    boolean[] row = new boolean[m];
    boolean[] col = new boolean[n];
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (matrix[i][j] == 0) {
                row[i] = col[j] = true;
            }
        }
    }
    for (int i = 0; i < m; i++) {
        for (int j = 0; j < n; j++) {
            if (row[i] || col[j]) {
                matrix[i][j] = 0;
            }
        }
    }
}

优化遍历方法：

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public void setZeroes(int[][] matrix) {
    int colLen = matrix.length, rowLen = matrix[0].length;
    boolean col0Flag = false, row0Flag = false;
    for (int i = 0; i < colLen; i++) {
        if (matrix[i][0] == 0) {
            col0Flag = true;
        }
    }
    for (int j = 0; j < rowLen; j++) {
        if (matrix[0][j] == 0) {
            row0Flag = true;
        }
    }
    for (int i = 1; i < colLen; i++) {
        for (int j = 1; j < rowLen; j++) {
            if (matrix[i][j] == 0) {
                matrix[i][0] = matrix[0][j] = 0;
            }
        }
    }
    for (int i = 1; i < colLen; i++) {
        for (int j = 1; j < rowLen; j++) {
            if (matrix[i][0] == 0 || matrix[0][j] == 0) {
                matrix[i][j] = 0;
            }
        }
    }
    if (col0Flag) {
        for (int i = 0; i < colLen; i++) {
            matrix[i][0] = 0;
        }
    }
    if (row0Flag) {
        for (int i = 0; i < rowLen; i++) {
            matrix[0][i] = 0;
        }
    }
}

📙1075. 项目员工 I

法一：通过聚合函数计算

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# Write your MySQL query statement below
select project_id,ROUND(AVG(experience_years), 2) average_years
from Project
left join Employee on Project.employee_id = Employee.employee_id
group by project_id;