class Solution {public:    int threeSumClosest(vector
     
       &num, int target) {        int len = num.size();        if (len < 3) {            return 0;        }        sort(num.begin(), num.end());                int sum = num[0] + num[1] + num[2];        for (int i=0; i
      
        target) {                    q--;                } else if (csum < target) {                    p++;                } else {                    return csum;                }            }        }        return sum;    }};
      
     

n方时间复杂度

 

第二轮：

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
     
       &num, int target) {        sort(num.begin(), num.end());        int len = num.size();        if (len < 3) {            return 0;        }        int min_diff = INT_MAX;        int closest = 0;        for (int i=0; i
      
        rest) {                    q--;                } else if (t < rest) {                    p++;                } else {                    break;                }            }        }        return closest;    }    };