848. Shifting Letters 题目链接:https://leetcode.com/problems/shifting-letters/description/ 思路:O(N^2)复杂度过不了:先处理shifts, 从尾到头执行shifts[i] = sum(shifts[i+1]+...+shifts[n-1]). 注意点:会爆int,及时取余或用longlong. 1 string shiftingLetters(string S, vector<int>& shift…

[LeetCode]848. Shifting Letters 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.me/ 题目地址:https://leetcode.com/problems/shifting-letters/description/ 题目描述: We have a string S of lowercase letters, and an integer arr…

原题链接在这里:https://leetcode.com/problems/shifting-letters/ 题目: We have a string S of lowercase letters, and an integer array shifts. Call the shift of a letter, the next letter in the alphabet, (wrapping around so that 'z' becomes 'a'). For example, shi…

问题描述: 问题规约为:对每一个数组S,移动(shifts[0] + shitfs[1]+...+shitfs[i] )mod 26位 def shiftingLetters(self, S: str, shifts: List[int]) -> str: #a 97 ord: char->int chr: int->char sm = sum(shifts) ans = '' for i in range(len(S)): ans += chr( (ord(S[i]) - 97 + s…

题目如下: 解题思路:本题首先要很快速的计算出任意一个字符shift后会变成哪个字符,其实也很简单,让shift = shift % 26,接下来再做计算.第二部是求出每个字符要shift的次数.可以得出S[0]的shift次数等于sum(shifts),S[1]的次数等于sum(shifts)-shifts[0],S[2]的次数等于sum(shifts)-shifts[0]-shifts[1]...依次类推.这个规律可以保证整个算法的复杂度是O(n). 代码如下: class Solution…

851. Loud and Rich 题目链接:https://leetcode.com/problems/loud-and-rich/description/ 思路:有向图DFS,记录最小的quiet值 注意点:可优化,记忆性搜索,每次搜到已经记录过的值时可直接比较不需要进一步搜下去. 1 void DFS(vector<int> &visited, vector<vector<int> >& G, int x,vector<int>&am…

849. Maximize Distance to Closest Person 题目链接:https://leetcode.com/problems/maximize-distance-to-closest-person/description/ 思路:pre[i]存放i之前离最近的1的距离.post记录之后的. res = max(min(pre[i],[post[i])) 注意点:初始nst需要设计极大或极小值. 1 int maxDistToClosest(vector<int>&am…

LeetCode Weekly Contest 23 1. Reverse String II Given a string and an integer k, you need to reverse the first k characters for every 2k characters counting from the start of the string. If there are less than k characters left, reverse all of them.…

LeetCode Weekly Contest 8 415. Add Strings User Accepted: 765 User Tried: 822 Total Accepted: 789 Total Submissions: 1844 Difficulty: Easy Given two non-negative numbers num1 and num2 represented as string, return the sum of num1 and num2. Note: The…

Weekly Contest 86 A:840. 矩阵中的幻方 3 x 3 的幻方是一个填充有从 1 到 9 的不同数字的 3 x 3 矩阵,其中每行,每列以及两条对角线上的各数之和都相等. 给定一个由整数组成的 N × N 矩阵,其中有多少个 3 × 3 的 “幻方” 子矩阵?(每个子矩阵都是连续的). 直接模拟即可,本来是签到题,由于粗心,浪费了时间. class Solution { public: int numMagicSquaresInside(vector<vector<int&…