题目描述

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).

思路

目的：找到三个数的和最接近target。

排序，从小到大，使与target之差值越来越小。i，j最小，k最大，然后去找合适值。

1. 如果三个数之和大于target，判断之差是否比目前最小差更小，更小就更新结果以及最小差，同时将j增大。
2. 反之同理。
3. 相等就是了。

代码

class Solution {public:int threeSumClosest(vector
     
      & nums, int target) {    if (nums.size() < 3)    {        return -1;    }    int res = 0;//最后答案    int distance = INT_MAX;//总的最近的差，包括大的和小的    int i, j, k;    sort(nums.begin(),nums.end());//先排序    for (i = 0; i < nums.size() - 2; i++)    {        j = i + 1;        k = nums.size() - 1;        while (j < k)        {            int temp = nums[i] + nums[j] + nums[k];            int temp_distance;            if (temp < target)//说明太小了，要变大            {                temp_distance = target - temp;//当前三个值与target的差                if (temp_distance < distance)//更接近了可以进行更新                {                    res = temp;                }                j++;            }            else if(temp > target)            {                    temp = nums[i] + nums[j] +     nums[k];                    temp_distance = temp - target;                if (temp_distance < distance)                {                res = temp;                }                k--;            }            else            {                temp = nums[i] + nums[j] + nums[k];                res = temp;            }        }    }    }};