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).

思路：这道题就是从一个数组中找到三个数的和与目标值最接近的值，然后返回。一般先对这个数组排序，时间复杂度在O(nlogn),然后在对这个排序后的数组处理，首先从头开始枚举一个数，然后再找出其后一个元素与最后一个元素，他们三个的数之和与目标值的差值，作为判断的依据，如果差值最小，则更新这个最小差值，如果这个差值大于0，则最后一个向前移动，相反，第二个向后移动。如此反复，就可以得出想要的目标值。

class Solution {

public:

    int threeSumClosest(vector<int> &num, int target) {

        int nSize=num.size();

        if(nSize<3)

            return 0;

        sort(num.begin(),num.end());

        int diff=0;

        int min_diff=num[0]+num[1]+num[2]-target;

        for(int i=0;i<=nSize-3;i++)

        {

            int j=i+1;

            int k=nSize-1;

            while(j<k)

            {

                diff=num[i]+num[j]+num[k]-target;

                if(diff==0)

                    return target;

                if(abs(diff)<abs(min_diff))

                    min_diff=diff;

                if(diff<0)

                    j++;

                else

                    k--;

            }

        }

        return target+min_diff;

    }

};