LeetCode之3SumClosest

题目：

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).

分析：题目大意是给定一个整型数组，要求求出其中三个数相加的结果与给定的目标值最接近。返回这个值。  还是和上一题差不多，重要的在于定界。

代码:

#include "stdafx.h"
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
class Solution{
public:
	int threeSumClosest(vector<int>& num, int target) {
		int min_gap = INT_MAX;
		int result = 0;
		sort(num.begin(),num.end());
		auto last = num.end();
		int sum,gap;
		for(auto i = num.begin();i < last-2;++i){
			auto j = next(i);
			auto k = prev(last);
			while(j<k){
				sum = *i + *j + *k;
				gap = abs(sum - target); //取绝对值
				if (min_gap > gap){
					min_gap = gap;
					result = sum;
				}

				if (sum > target)
					--k;
				else
					++j;
			}			
		}
        return result;
	}
};

int _tmain(int argc, _TCHAR* argv[])
{
	vector<int> v1;
	int rtn;
	v1.push_back(-1);
	v1.push_back(1);
	v1.push_back(2);
	v1.push_back(-4);

	Solution s;
	rtn = s.threeSumClosest(v1, 1);
	cout << rtn << endl;
	
	system("pause");
	return 0;
}