标签：style blog io ar color os sp for on
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)
标签：style blog io ar color os sp for on
原文地址：http://www.cnblogs.com/55open/p/4141068.html
