leetcode 16 3Sum Closest

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<int>& 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;
			}
		}
	}
	}
};