标签：while exactly 细节 pre div solution min 算法 else
算法描述：
Given an array nums of n integers and an integer target, find three integers in nums such that the sum is closest to target. Return the sum of the three integers. You may assume that each input would have exactly one solution.
Example:
Given array nums = [-1, 2, 1, -4], and target = 1. The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
解题思路：排序，注意细节
int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(),nums.end()); int minGap = INT_MAX; int res = INT_MAX; for(int i=0; i < nums.size(); i++){ int low = i+1; int high = nums.size()-1; while(low < high ){ int sum = nums[i] + nums[low] + nums[high]; int gap = abs(target - sum); if(minGap > gap){ minGap = gap; res = sum; } if(target < sum) high--; else low++; } } return res; }
标签：while exactly 细节 pre div solution min 算法 else
原文地址：https://www.cnblogs.com/nobodywang/p/10326780.html
