LeetCode 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).

class Solution {
public:
    int threeSumClosest(vector<int> &num, int target) {
        sort(num.begin(),num.end());
       int min = INT_MAX;
       int len = num.size();
       int sum = 0;
       int result = 0;
       for(int i=0;i<len-2;i++)
       {
           int j = i + 1;
           int k = len - 1;
           while(j<k)
           {
               sum = num[i] + num[j] + num[k];
               if(abs(sum-target) < min)
                {
                    min = abs(sum - target);
                    result = num[i] + num[j] + num[k];
                }
                if(sum < target)j++;
                if(sum > target)k--;
                if(sum == target)
                    return result;
           }
       }
       return result;
    }
};

解题思路：最直观的想法，三次循环，时间复杂度o(n^3),显然不能接受。

　　　　想到剪枝，在2次循环时，有很多情况是不必要走的，即如果得知num[i] + num[j] + num[k] > target的情况下，表明num[i] + num[m] + num[k] > target，其中m > j，同理还有小于的情况，于是复杂度可以降为o(n^2)