leetcode 16 -- 3Sum Closest

3Sum Closest

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).

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题意：
给定一个数组和一个目标值，返回其中3个数字的和且和值最接近目标值。

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思路：
先把数组排序，把3个数字的问题转换为找2个数字问题，每次固定一个数值，那么target就变为了（target-固定的数字），找两个数字近似新target就简单多了，定义两个指针，指向首和尾，定义一个Min，表示与新target的差值，差值越小越接近，不断的向后移动首指针或者向前移动尾指针来调整近似度Min，直到轮寻完毕为止，也就是每个值都被固定过。

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代码：

class Solution {public:    int threeSumClosest(vector<int>& nums, int target) {        if(nums.size() == 0){            return 0;        }        //排序        sort(nums.begin(), nums.end());        auto End = nums.end()-1;        //初始化Min，为了第一次可以更新        int Min = 2147483647;        int ret = 0;        //每次固定一个值，轮寻完毕为止，注意最小的为nums.end()-2，因为每次是3个值来判断。        for(auto iter = nums.begin(); iter != nums.end()-1; ++iter){            //第一个迭代器为首，第二个为尾            auto iter1 = iter + 1;            auto iter2 = nums.end()-1;            //减去固定值产生新的target            int my_target = target-*iter;            int tmp;            //首尾迭代器不重合            while(iter1 != iter2){                tmp = *iter1 + *iter2;                if(tmp < my_target){                    //判断是否有更小的近似度Min                    if(abs(my_target-tmp) < Min){                        Min = abs(my_target-tmp);                        ret = *iter1 + *iter2 + *iter;                    }                    //移动迭代器                    ++iter1;                }                else if(tmp > my_target){                    if(abs(my_target-tmp) < Min){                        Min = abs(my_target-tmp);                        ret = *iter1 + *iter2 + *iter;                    }                    --iter2;                }else{                    //如果相等说明找到了等于target的三个值                    return target;                }            }        }        return ret;    }};