Leetcode 16. 3Sum Closest
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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).
题目大意:取三个数,求出与给的目标值最接近的三个数的和。
题目分析:主要思路与第15题差不多,只需要在其原有代码基础上稍作修改就行
代码1:运行时间12ms
class Solution {public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end());//先排序 int i,j,k,ans,sum,differ=INT_MAX,tempdiff; for(i=0;i<nums.size()-1;i++){ if(i>0&&nums[i-1]==nums[i])//去除重复 continue; j=i+1;k=nums.size()-1; while(j<k){ sum=nums[i]+nums[j]+nums[k]; tempdiff=abs(sum-target);//记录当前差 if(tempdiff<differ){ ans=sum; differ=tempdiff;//differ记录最小差 } if(sum==target) return target; else if(sum<target){ while(j++<k&&nums[j-1]==nums[j]);//去除重复 }else{ while(j<--k&&nums[k]==nums[k+1]);//去除重复 } } } return ans; }};代码2:下面试discuss里面的代码,思路一样的,写的更加简便易懂,不过运行时间稍长 28ms
class Solution {public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int res = nums[0] + nums[1] + nums[2]; for(int i = 0; i < nums.size() - 2; i++){ int j = i + 1, k = nums.size() - 1; while(j < k){ int num = nums[i] + nums[j] + nums[k]; if(abs(num - target) < abs(res - target)) res = num; if(num < target) j++; else k--; } } return res; }};
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