#Leetcode# 16. 3Sum Closest
https://leetcode.com/problems/3sum-closest/
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).
代码1:
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { int n = nums.size(); sort(nums.begin(), nums.end()); int out = nums[0] + nums[1] + nums[2]; int minn = abs(target - out); for(int i = 0; i < n - 2; i ++) { int l = i + 1, r = n - 1; while(l < r) { int num = nums[i] + nums[l] + nums[r]; if(minn > abs(num - target)) { minn = abs(num - target); out = num; } if(num > target) r --; else l ++; } } return out; } };
这个是之前 WA 的一份改过来的虽然不知道之前的错在哪里但是 AC 了就好了!
代码2:
class Solution { public: int threeSumClosest(vector<int>& nums, int target) { sort(nums.begin(), nums.end()); int n = nums.size(); //int out = nums[0] + nums[1] + nums[2]; //int minn = abs(out - target); int out = 0; int minn = INT_MAX; for(int i = 0; i < n - 2; i ++) { int l = i + 1, r = n - 1; while(l < r) { int num = nums[i] + nums[l] + nums[r]; if(abs(target - num) < minn) { minn = abs(target - num); out = num; } if(num > target) r --; else l ++; } } return out; } };
可能是在拉低通过率吧。。。
这个题目和上一个 $3Sum$ 差不多的 都是 $O(n^2)$ 的时间复杂度 因为各种小细节的错误 WA 了很多次 但是用和上一个一样的写法写错了 没找出来哪里有问题 改好之后再贴出来 嘻嘻
☺好像变成了一个爱碎碎念的 be 主