LeetCode刷题 | Brust’Blog

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Dec 13, 2022

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leetcode

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记录刷题代码

LeetCode-1. 两数之和

code1


class Solution {
    public int[] twoSum(int[] nums, int target) {
        //nums = [2, 7, 11, 15], target = 9
        //nums = [2, 7, 11, 15], target = 9
        for (int i = 0; i < nums.length; i++) {
            for (int j = i+1; j < nums.length; j++) {
                if(nums[i]+nums[j] == target){
                    return new int[]{i,j};
                }
            }
        }
        return null;
    }
}

code2


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

LeetCode-2. 两数相加

code


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int n = 0;
        ListNode rs = new ListNode();
        ListNode node = rs;
        ListNode q = l1;
        ListNode p = l2;
        while (q != null || p != null) {
            int i = 0;
            int j = 0;
            if (q != null) {
                i = q.val;
            }
            if (p != null) {
                j = p.val;
            }
            int num = i + j + n;
            n = num / 10;
            node.next = new ListNode(num % 10);
            node = node.next;
            if (q != null) {
                q = q.next;
            }
            if (p != null) {
                p = p.next;
            }
        }
        if (n > 0) {
            node.next = new ListNode(n);
        }
        return rs.next;
    }
}

LeetCode-3. 无重复字符的最长子串

code


class Solution {
    public int lengthOfLongestSubstring(String s) {
        // 哈希集合，记录每个字符是否出现过
        Set<Character> occ = new HashSet<Character>();
        int n = s.length();
        // 右指针，初始值为 -1，相当于我们在字符串的左边界的左侧，还没有开始移动
        int rk = -1, ans = 0;
        for (int i = 0; i < n; ++i) {
            if (i != 0) {
                // 左指针向右移动一格，移除一个字符
                occ.remove(s.charAt(i - 1));
            }
            while (rk + 1 < n && !occ.contains(s.charAt(rk + 1))) {
                // 不断地移动右指针
                occ.add(s.charAt(rk + 1));
                ++rk;
            }
            // 第 i 到 rk 个字符是一个极长的无重复字符子串
            ans = Math.max(ans, rk - i + 1);
        }
        return ans;
    }
}

LeetCode-4. 寻找两个正序数组的中位数

code


class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int[] arr = new int[nums1.length + nums2.length];
        int index = 0;
        double rs = 0;
        for (int i : nums1) {
            arr[index++] = i;
        }
        for (int i : nums2) {
            arr[index++] = i;
        }
        Arrays.sort(arr);
        if (arr.length % 2 == 0) {
            int i = arr.length / 2;
            int j = i - 1;
            rs = (arr[i] + arr[j]) / 2.0;
        } else {
            int i = arr.length / 2;
            rs = arr[i];
        }
        return rs;
    }
}

LeetCode-5. 最长回文子串

code


class Solution {
    public String longestPalindrome(String s) {
        int n = s.length();
        String res = "";
        boolean[][] dp = new boolean[n][n];
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i; j < n; j++) {
                dp[i][j] = s.charAt(i) == s.charAt(j) && (j - i < 2 || dp[i + 1][j - 1]); //j - i 代表长度减去 1
                if (dp[i][j] &&  j - i + 1 > res.length()) {
                    res = s.substring(i, j + 1);
                }
            }
        }
        return res;
    }
}

LeetCode-9. 回文数

code


class Solution {
    public boolean isPalindrome(int x) {
        if( x < 0){
            return false;
        }
        int div = 1;
        while(x / div >= 10){
            div *= 10;
        }
        while(x > 0){
            int left = x/div;
            int right = x%10;
            if(left != right){
                return false;
            }
            x = (x % div)/10;
            div /=100;
        }
        return true;
    }
}

LeetCode-15. 三数之和

code


class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        Arrays.sort(nums);
        List<List<Integer>> res = new ArrayList<>();
        for(int k = 0; k < nums.length - 2; k++){
            if(nums[k] > 0) break;
            if(k > 0 && nums[k] == nums[k - 1]) continue;
            int i = k + 1, j = nums.length - 1;
            while(i < j){
                int sum = nums[k] + nums[i] + nums[j];
                if(sum < 0){
                    while(i < j && nums[i] == nums[++i]);
                } else if (sum > 0) {
                    while(i < j && nums[j] == nums[--j]);
                } else {
                    res.add(new ArrayList<Integer>(Arrays.asList(nums[k], nums[i], nums[j])));
                    while(i < j && nums[i] == nums[++i]);
                    while(i < j && nums[j] == nums[--j]);
                }
            }
        }
        return res;
    }
}

LeetCode-175. 组合两个表

code


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

LeetCode-180. 连续出现的数字

code


select distinct Num as ConsecutiveNums
from (
  select Num, 
    case 
      when @prev = Num then @count := @count + 1
      when (@prev := Num) is not null then @count := 1
    end as CNT
  from Logs, (select @prev := null,@count := null) as t
) as temp
where temp.CNT >= 3

LeetCode-181. 超过经理收入的员工

code


# Write your MySQL query statement below
select e.Name as Employee
from Employee e
left join Employee f on e.ManagerId = f.Id
where e.Salary > f.Salary

LeetCode-182. 查找重复的电子邮箱

code


# Write your MySQL query statement below
select Email
from Person
group by Email
having count(Email) >1

LeetCode-183. 从不订购的客户

code


# Write your MySQL query statement below
select c.Name as Customers
from Customers c
left join Orders o on c.Id = o.CustomerId
where o.id is null