942. 增减字符串匹配 942. DI String Match 题目描述 每日一算法2019/6/21Day 49LeetCode942. DI String Match Java 实现 and so on 参考资料 https://leetcode-cn.com/problems/di-string-match/ https://leetcode.com/problems/di-string-match/…

Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I", then A[i] < A[i+1] If S[i] == "D&qu…

这是悦乐书的第361次更新,第388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223题(顺位题号是942).给定仅包含I(增加)或D(减少)的字符串S,令N = S.length. 返回元素值范围为[0,1,-,N]的整型数组A,使得对于所有i = 0,-,N-1: 如果S[i] =='I',那么A[i] < A[i + 1]. 如果S[i] =='D',那么A[i] > A[i + 1]. 例如: 输入:"IDID" 输出:[0,4,1…

686. 重复叠加字符串匹配 686. Repeated String Match 题目描述 给定两个字符串 A 和 B,寻找重复叠加字符串 A 的最小次数,使得字符串 B 成为叠加后的字符串 A 的子串,如果不存在则返回 -1. 举个例子,A = "abcd",B = "cdabcdab". 答案为 3,因为 A 重复叠加三遍后为 "abcdabcdabcd",此时 B 是其子串:A 重复叠加两遍后为 "abcdabcd",…

Given two strings A and B, find the minimum number of times A has to be repeated such that B is a substring of it. If no such solution, return -1. For example, with A = "abcd" and B = "cdabcdab". Return 3, because by repeating A three…

problem 942. DI String Match 参考 1. Leetcode_easy_942. DI String Match; 完…

https://leetcode.com/problems/di-string-match/ Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I&quo…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/di-string-match/description/ 题目描述 Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. R…

题目要求 Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I", then A[i] < A[i+1] If S[i] == "…

题目如下: Given a string S that only contains "I" (increase) or "D" (decrease), let N = S.length. Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1: If S[i] == "I", then A[i] < A[i+1] If S[i] == &quo…