首页 > 代码库 > 每日算法之十五:threesumClosset
每日算法之十五:threesumClosset
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).
求解跟目标数值最近的三元组。<span style="font-size:14px;">class Solution { public: int threeSumClosest(vector<int> &num, int target) { sort(num.begin(), num.end()); int ret; bool first = true; for(int i = 0; i < num.size(); i++) { int j = i + 1; int k = num.size() - 1; while(j < k) { int sum = num[i] + num[j] + num[k]; if (first) { ret = sum; first = false; } else { if (abs(sum - target) < abs(ret - target)) ret = sum; } if (ret == target) return ret; if (sum > target) k--; else j++; } } return ret; } }; </span>
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。