3Sum Closest

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).

类似于3Sum的处理, 在处理3Sum的时候，是将所有满足条件的和都保存起来，在处理此问题的时候，则要求解每次twoSum中与curtarget差值最小的和，最后在外层循环再求与最终target差值最小的和

class Solution {public:    //类似于3Sum的处理, 在处理3Sum的时候，是将所有满足条件的和都保存起来，在处理此问题的时候，    //则要求解每次twoSum中与curtarget差值最小的和，最后在外层循环再求与最终target差值最小的和    int twoSum(vector<int> &nums, int n, int curStart, int curTarget)    {        int low = curStart, high = n - 1;        int minDiff = 99999999;        while(low < high)        {            int tmpSum = nums[low] + nums[high];            if( tmpSum == curTarget)            {                return tmpSum;            }            else if(tmpSum < curTarget)            {                int tmps = nums[low];                while(low < high && tmps == nums[low])                    ++low;            }            else            {                int tmps = nums[high];                while(low < high && tmps == nums[high])                    --high;            }                        minDiff = abs(tmpSum - curTarget) < abs(minDiff - curTarget) ? tmpSum : minDiff;        }                return minDiff;    }        int threeSumClosest(vector<int>& nums, int target) {        int len = nums.size();        if(len < 3)            return 0;        sort(nums.begin(), nums.end());        int minDiff = 99999999, i = 0;        while(i < len)        {            int curTarget = target - nums[i];            int diff = twoSum(nums, len, i + 1, curTarget);            minDiff = abs(minDiff - target) > abs(diff + nums[i] - target) ? diff + nums[i] : minDiff;            int tmps = nums[i];            while(i < len && nums[i] == tmps)                ++i;        }                return minDiff;    }};