Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

题意: 判断两个字符串是否互为Scramble字符串,而互为Scramble字符串的定义: 字符串看作是父节点,从字符串某一处切开,生成的两个子串分别是父串的左右子树,再对切开生成的两个子串继续切开,直到无法再切,此时生成为一棵二叉树.对二叉树的任一子树可任意交换其左右分支,如果S1可以通过交换变成S2,则S1,S2互为Scramble字符串. 思路: 对于分割后的子串,应有IsScramble(s1[0,i] , s2[0,i]) && IsSCramble(s1[i,length] ,…

题目: Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we ma…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

原题地址 两个字符串满足什么条件才称得上是scramble的呢? 如果s1和s2的长度等于1,显然只有s1=s2时才是scramble关系. 如果s1和s2的长度大于1,那么就对s1和s2进行分割,划分成两个子问题分别处理. 如何分割呢?当然不能任意分割.假设分割后s1变成了s11和s12,s2变成了s21和s22,那么只有2种分割方式: 1. s11.length = s21.length & s12.length = s22.length,如下图所示: s1: ? ? ? ? ? -----…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

题目: Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we ma…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

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 递归 动态规划 日期 题目地址:https://leetcode.com/problems/scramble-string/description/ 题目描述 Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty subs…

转:http://www.cnblogs.com/easonliu/p/3696135.html 分析:这个问题是google的面试题.由于一个字符串有很多种二叉表示法,貌似很难判断两个字符串是否可以做这样的变换.对付复杂问题的方法是从简单的特例来思考,从而找出规律.先考察简单情况:字符串长度为1:很明显,两个字符串必须完全相同才可以.字符串长度为2:当s1="ab", s2只有"ab"或者"ba"才可以.对于任意长度的字符串,我们可以把字符串s…

Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…

Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…

Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…

Scramble String Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the s…

错误提示: Caused by: org.apache.ibatis.executor.ExecutorException: No constructor found in com.tuyrk._161_java_socket.project6.entity.User matching [java.lang.Long, java.lang.String, java.lang.String] 这里就有点坑了,明明提示的是没有三个参数的构造函数,然而我添加上三个参数的构造函数还是报错 解决方法: 添…

第87节:Java中的Bootstrap基础与SQL入门 前言复习 什么是JQ? : write less do more 写更少的代码,做更多的事 找出所有兄弟: $("div").siblings() 基本过滤器: 选择器:过滤器 $("div:first") :first: 找到第一个元素 :last: 找到最后一个元素 :even: 找出偶数索引 :odd: 找出奇叔索引 :gt(index): 大于 :lt(index): 小于 :eq(index): 等…

java.lang.String & java.lang.StringBuilder String 成员方法 作用 public charAr(int index) 返回给定位置的代码单元 public codePointAt(int index) 返回给定位置的码点 int compareTo(String other) 根据字典序,如果this在other前,返回正数,在后返回负数,相等返回0;和常量比较时,应该把常量写在前面,防止空指针异常 boolean equals(String ot…

leetcode87. Scramble String 题意: 给定一个字符串s1,我们可以通过将它分解为两个非空子字符串来表示为二叉树. 思路: 递归解法 对于每对s1,s2. 在s1某处切一刀,s1分成left,right,然后在s2首部开始等长的地方切一刀,切成left,right.只要s1的left和s2的left,s1的right和s2的right同样也构成scramble string就可以了. 递归解法需要剪枝. 剪枝1:s1,s2不等长return 剪枝2:s1,s2字符不相同r…

(Version 0.0) 作为一个小弱,这个题目是我第一次碰到三维的动态规划.在自己做的时候意识到了所谓的scramble实际上有两种可能的类型,一类是在较低层的节点进行的两个子节点的对调,这样的情况如果我们从第一层切分点,或者说从较高层的切分点看的话,s1和s2切分点左边的子串所包含的字符的种类个数应该完全一致,同样右边也是完全一致:另一类是在较高层切分点进行的互换,这样我们如果在同层来考察s1和s2的话,会发现s1的切分点左侧的char和s2的切分点右侧的char种类和每种char的数目一…

org.springframework.boot.context.properties.bind.BindException: Failed to bind properties under 'logging.level' to java.util.Map<java.lang.String, java.lang.String> at org.springframework.boot.context.properties.bind.Binder.handleBindError(Binder.ja…

java.lang.IllegalStateException: Failed to load ApplicationContext at org.springframework.test.context.cache.DefaultCacheAwareContextLoaderDelegate.loadContext(DefaultCacheAwareContextLoaderDelegate.java:125) at org.springframework.test.context.suppo…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": To scramble the string, we may choose any non-leaf node and swap its two ch…

题目: Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we ma…

Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. Below is one possible representation of s1 = "great": great / \ gr eat / \ / \ g r e at / \ a t To scramble the string, we may ch…