Problem: Solve a given equation and return the value of x in the form of string “x=#value”. The equation contains only ‘+’, ‘-’ operation, the variable x and its coefficient. If there is no solution for the equation, return “No solution”. If there are infinite solutions for ...
代码:(比较函数用了C++11中的lambda表达式) 1classSolution {2public:3vector<pair<int,int>> reconstructQueue(vector<pair<int,int>>&people) {4vector<pair<int,int>>result;5sort(people.begin(), people.end(), [](constpair<int,int>& p1,constpair<int,int>&p2)6{7if(p1.first ==p2.first)...
AI代码解释 classSolution:deffindOriginalArray(self,changed:List[int])->List[int]:cnt=Counter(changed)keys=sorted(cnt.keys())res=[]forkinkeys:ifk==0:ifcnt[k]%2!=0:return[]res.extend([k]*(cnt[k]// 2))continueifcnt[k]==0:continueifcnt[2*k]<cnt[k]:return[]res.extend([k]*cnt...
代码: 1classSolution {2public:3boolcanPartition(vector<int>&nums) {4intsum =0;5for(inti =0; i < nums.size(); ++i) {6sum +=nums[i];7}8if(sum %2==1) {9returnfalse;10}11sum /=2;12intdp[nums.size()][sum] = {false};13if(nums.size() ==1) {14returnfalse;15}16for(...
https://leetcode.cn/contest/weekly-contest-400/problems/minimum-number-of-chairs-in-a-waiting-room/ 题意 直接枚举 思路 统计当前字符串中最大的E的统计数目即可,与多任务CPU类似。 复杂度分析: 时间复杂度:O(n \log n), 其中n表示数组的长度。
class Solution { public int minOperations(int[] nums1, int[] nums2) { // 无解情况判断 if (nums1.length > nums2.length * 6 || nums2.length > nums1.length * 6) { return -1; } // 保证 nums1.sum > nums2.sum if (Arrays.stream(nums1).sum() < Arrays.stream(nums2).sum()...
https://leetcode-cn.com/contest/weekly-contest-174/ranking/1/ 总体心得 使用python3,做成了第一道题,第二道题的解超时,便结束了第一次周赛。 下午看了排行榜靠前的参赛者代码,最快的在17分钟内就完赛了(使用python3),使用python3的和c++的数量几乎对半分。
Algorithms Exercise: LeetCode Problems, LeetCode Weekly Contest etc. github.com/caipengbo/LeetCode-CPP Resources Readme Activity Stars 56 stars Watchers 6 watching Forks 17 forks Report repository Releases No releases published Packages No packages published Languages Java 99.9% C++ 0.1...
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}classSolution {public:intshipWithinDays(vector<int>& weights,intD) {intnum[50100] = {0};intcnt =0;intlen =weights.size();for(inti =0; i < len ; i++){ num[i]=weights[i]; }returnBinarySearch(num ,len, D); } }; 1. ...