LeetCode之16_3Sum Closest

题目原文：

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).

题意分析：

与题号为15的3sum相似，只不过这道题要求三个数的和是给定的一个数字，即找到a+b+c=target，解题思路相同，将两个数的求和目标变为target-a即可。

解题代码：

//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).

//从给出的数组中找出三个数字，使其和与给定的目标最为接近

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

class Solution {
public:
int threeSumClosest(vector<int>& nums, int target) {
if (nums.size()<3)
{
return -1;
}
int TarSum = 0;
int tempSum = 0;
int nMin = 1024;
int nRet =0;

sort(nums.begin(),nums.end()); //使容器中的数字为有序序列

for (int i=0; i<nums.size()-2; i++)
{
TarSum = target-nums[i]; //声明搜索的目标和 a+b+c = target, a+b = target-c
int j = i+1; //声明搜索的首尾节点
int k=nums.size()-1;

while(j<k)
{
tempSum = nums[j]+nums[k]-TarSum;

if (abs(tempSum) < nMin)
{
nMin = abs(tempSum);
nRet = nums[i]+nums[j]+nums[k];
}

if (tempSum < 0)
{
j++;
}
else if (tempSum > 0)
{
k--;
}
else if (tempSum == 0)
{
return target;
}
}
}

return nRet;
}
};