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;
    }
};