#Leetcode# 16. 3Sum Closest

https://leetcode.com/problems/3sum-closest/

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

Example:

Given array nums = [-1, 2, 1, -4], and target = 1.

The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

代码1：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        int out = nums[0] + nums[1] + nums[2];
        int minn = abs(target - out);
        
        for(int i = 0; i < n - 2; i ++) {
            int l = i + 1, r = n - 1;
            while(l < r) {
                int num = nums[i] + nums[l] + nums[r];
                if(minn > abs(num - target)) {
                    minn = abs(num - target);
                    out = num;
                }
                
                if(num > target) r --;
                else l ++;
            }
        }
        return out;
    }
};

这个是之前 WA 的一份改过来的虽然不知道之前的错在哪里但是 AC 了就好了！

代码2：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        //int out = nums[0] + nums[1] + nums[2];
        //int minn = abs(out - target);
        int out = 0;
        int minn = INT_MAX;
        for(int i = 0; i < n - 2; i ++) {
            int l = i + 1, r = n - 1;
            while(l < r) {
                int num = nums[i] + nums[l] + nums[r];
                if(abs(target - num) < minn) {
                    minn = abs(target - num);
                    out = num;
                }
                if(num > target) r --;
                else l ++;
            }
        }
        return out;
    }
};

　　

 可能是在拉低通过率吧。。。

这个题目和上一个 $3Sum$ 差不多的 都是 $O(n^2)$ 的时间复杂度 因为各种小细节的错误 WA 了很多次 但是用和上一个一样的写法写错了 没找出来哪里有问题 改好之后再贴出来 嘻嘻

☺好像变成了一个爱碎碎念的 be 主