[LeetCode]3Sum Closest题解

3sum Closest:

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

For example, given array S = {-1 2 1 -4}, and target = 1.

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

这道题的解题思路是比较常规的。首先将数组排序，设置三个下标first、second、third分别初始化为0，1，nums.size()-1。

1、如果这三个下标对应的数据之和大于target，那么就把third减一，

2、如果小于target，就把second加一。

3、如果等于target，结束查找，返回这个值。

每当second和third相遇的时候就让first+1，重置second = first + 1，thrid = nums.size() - 1。

这样就可以保证扫描到所有可能的结果，并且算法复杂度为O(n^2).

代码如下：

class Solution {
public:
    int threeSumClosest(vector<int>& nums, int target) {
        if(nums.size()<3) return 0;
        int first = 0, second = 1, third = nums.size() - 1;
        sort(nums.begin(),nums.end());
        int re = nums[0] + nums[1] + nums[2];
        for(;first < nums.size()-2;first++){
            second = first + 1;
            third = nums.size() - 1;
            while(second < third){
                int temp = nums[first] + nums[second] + nums[third];
                if(abs(re - target) > abs(temp - target)) re = temp;
                if(temp == target){
                    return temp;
                }
                else if(temp < target){
                    second ++;
                }
                else{
                    third --;
                }
            }
        }
        return re;
    }
};