leetcode-16:3sum closest最近的三数之和

题目：

Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution. Example: Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

给定一个包括 n 个整数的数组 nums 和 一个目标值 target。找出 nums 中的三个整数，使得它们的和与 target 最接近。返回这三个数的和。假定每组输入只存在唯一答案。 例如，给定数组 nums = [-1，2，1，-4], 和 target = 1. 与 target 最接近的三个数的和为 2. (-1 + 2 + 1 = 2).

思路：和上一题类似，要保证和target的差 最小，那么定义 一个变量diff保存他们之间的差。首先排序，然后固定一个位置，剩下的两个数left从左往右，right从又往左，每确定两个数，我们求出此三数之和，然后算和给定值的差的绝对值存在newDiff中，然后和diff比较并更新diff和结果closest即可

class Solution {
public:
    int threeSumClosest(vector& nums, int target) {
        int closest = nums[0] + nums[1] + nums[2];
        int diff = abs(closest - target);
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size() - 2; ++i) {
            int left = i + 1, right = nums.size() - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                int newDiff = abs(sum - target);
                if (diff > newDiff) {
                    diff = newDiff;
                    closest = sum;
                }
                if (sum < target) ++left;
                else --right;
            }
        }
        return closest;
    }
};

参考：http://www.cnblogs.com/grandyang/p/4510984.html