problem 888. Fair Candy Swap solution: class Solution { public: vector<int> fairCandySwap(vector<int>& A, vector<int>& B) { , sumB = , delta = ; unordered_set<int> setA; for(auto a:A) { sumA += a; setA.insert(a);//err... }…

作者: 负雪明烛 id: fuxuemingzhu 个人公众号: 每日算法题 本文关键词:力扣,LeetCode,算法题,算法,Python 目录 题目描述 题目大意 解题方法 代码 刷题心得 关于作者 日期 题目地址:https://leetcode.com/problems/fair-candy-swap/description/ 题目描述 Alice and Bob have candy bars of different sizes: A[i] is the size of the i-…

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so that aft…

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so that aft…

题目要求 Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so tha…

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so that aft…

题目: Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so that…

这是悦乐书的第339次更新,第363篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第208题(顺位题号是888).Alice和Bob有不同大小的糖果棒:A[i]是Alice拥有的第i个糖果棒的大小,B[j]是Bob拥有的第j个糖果棒的大小. 由于他们是朋友,他们想交换一个糖果,以便交换后,他们都有相同的糖果总量. (一个人拥有的糖果总量是他们拥有的糖果大小的总和.) 返回一个整数数组ans,其中ans[0]是Alice必须交换的糖果的大小,ans[1]是Bob必须交…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/3758 访问. 爱丽丝和鲍勃有不同大小的糖果棒:A[i] 是爱丽丝拥有的第 i 块糖的大小,B[j] 是鲍勃拥有的第 j 块糖的大小. 因为他们是朋友,所以他们想交换一个糖果棒,这样交换后,他们都有相同的糖果总量.(一个人拥有的糖果总量是他们拥有的糖果棒大小的总和.) 返回一个整数数组 ans,其中 ans[0] 是爱丽丝必须交换的糖果棒的大小,ans[1] 是…

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so that aft…

Alice and Bob have candy bars of different sizes: A[i] is the size of the i-th bar of candy that Alice has, and B[j] is the size of the j-th bar of candy that Bob has. Since they are friends, they would like to exchange one candy bar each so that aft…

如何在红帽 企业版Linux系统中添加swap文件? 解决方法: 1. 确定swap文件的大小,单位为M.将该值乘以1024得到块大小.例如,64MB的swap文件的块大小是65536. 2. 在root提示符下,键入下面的命令,其中count大小等于所要求的块大小: dd if=/dev/zero f=/swapfile bs=1024 count=65536 3. 建立swap文件: mkswap /swapfile 4. 要立即打开swap文件而不是在启动时自动开启: swapon /sw…

problem 905. Sort Array By Parity solution1: class Solution { public: vector<int> sortArrayByParity(vector<int>& A) { vector<int> even, odd; for(auto a:A) { ==) even.push_back(a); else odd.push_back(a); } even.insert(even.end(), odd.…

problem 922. Sort Array By Parity II solution1: class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { int n = A.size(); , j=; i<n, j<n; ) { ==) i += ; ==) j += ; if(j<n) swap(A[i], A[j]); } return A; } }; 参考 1…

problem 599. Minimum Index Sum of Two Lists 题意:给出两个字符串数组,找到坐标位置之和最小的相同的字符串. 计算两个的坐标之和,如果与最小坐标和sum相同,那么将这个字符串加入结果res中,如果比sum小,那么sum更新为这个较小值,然后将结果res清空并加入这个字符串. solution: class Solution { public: vector<string> findRestaurant(vector<string>&…

Description 有 \(k\) 个长度为 \(n\) 的只含 \(a\) 或 \(b\) 字符串,并不知道它们具体是多少,只知道它们的字典序不小于字符串 \(A\),同时不大于字符串 \(B\).定义一个字符串是合法的当且仅当它是这 \(k\) 个字符串之一的前缀(如果它是多个串的前缀那么只计算一次).求合法的字符串最大可能是多少 Input 第一行是两个整数 \(n\) 和 \(k\) 下面两行,第一行是长度为 \(n\) 的字符串 \(A\),第二行是长度为 \(n\) 的字符串 \…

Description 给定 \(n\) 个商店,他们围成一个圆圈,按照顺时针从 \(1\) 到 \(n\) 编号.你有 \(T\) 元钱,从 \(1\) 号点开始按照顺时针方向走,每到一个商店,只要钱够就必须买这个商店的物品.商店中物品是无限的,即多次到达可能多次购买.求会买多少件物品 Input 第一行是一个整数 \(n\) 下面一行 \(n\) 个整数 \(a_i\),代表每个商店物品的价格 Output 一行一个整数代表答案 Hint \(1~\leq~n~\leq~2~\times~1…

题意:初始有t元,每次从1开始买,从1到n依次有n个人,每个人的东西价格为a[i],该人依次能买就买,到n之后再回到1从头开始,问最后能买到的东西数量 n<=2e5,t<=1e18,a[i]<=1e9 思路:显然购买是有周期的,每次周期变化都会少至少一个人,所以至多进行n次周期的变化 #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<…

problem 1021. Remove Outermost Parentheses 参考 1. Leetcode_easy_1021. Remove Outermost Parentheses; 完…

problem 1022. Sum of Root To Leaf Binary Numbers 参考 1. Leetcode_easy_1022. Sum of Root To Leaf Binary Numbers; 完…

problem 1025. Divisor Game 参考 1. Leetcode_easy_1025. Divisor Game; 完…

problem 1029. Two City Scheduling 参考 1. Leetcode_easy_1029. Two City Scheduling; 完…

problem 1030. Matrix Cells in Distance Order 参考 1. Leetcode_easy_1030. Matrix Cells in Distance Order; 完…

problem 1033. Moving Stones Until Consecutive 参考 1. Leetcode_easy_1033. Moving Stones Until Consecutive; 完…

problem 1037. Valid Boomerang 参考 1. Leetcode_easy_1037. Valid Boomerang; 完…

problem 1042. Flower Planting With No Adjacent 参考 1. Leetcode_easy_1042. Flower Planting With No Adjacent; 完…

problem 1046. Last Stone Weight 参考 1. Leetcode_easy_1046. Last Stone Weight; 完…

problem 1047. Remove All Adjacent Duplicates In String 参考 1. Leetcode_easy_1047. Remove All Adjacent Duplicates In String; 完…

problem 1051. Height Checker solution class Solution { public: int heightChecker(vector<int>& heights) { vector<int> orders = heights; sort(orders.begin(), orders.end()); ; ; i<heights.size(); i++) { if(heights[i]!=orders[i]) res++; } r…

problem 1071. Greatest Common Divisor of Strings solution class Solution { public: string gcdOfStrings(string str1, string str2) { return (str1+str2==str2+str1) ? (str1.substr(, gcd(str1.size(), str2.size()))) : ""; } }; 参考 1. Leetcode_easy_1071…