给定一个字符串 s1,我们可以把它递归地分割成两个非空子字符串,从而将其表示为二叉树.下图是字符串s1 = "great"的一种可能的表示形式. great / \ gr eat / \ / \g r e at / \ a t在扰乱这个字符串的过程中,我们可以挑选任何一个非叶节点,然后交换它的两个子节点.例如:如果我们挑选非叶节点 "gr",交换它的两个子节点,将会产生扰乱字符串&q…
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…
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": To scramble the string, we may choose any non-leaf node and swap its two 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…
Hard! 题目描述: 给定一个字符串 s1,我们可以把它递归地分割成两个非空子字符串,从而将其表示为二叉树. 下图是字符串 s1 = "great" 的一种可能的表示形式. great / \ gr eat / \ / \ g r e at / \ a t 在扰乱这个字符串的过程中,我们可以挑选任何一个非叶节点,然后交换它的两个子节点. 例如,如果我们挑选非叶节点 "gr" ,交换它的两个子节点,将会产生扰乱字符串 "rgeat" . rgea…
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…
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…
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…
87. 扰乱字符串 给定一个字符串 s1,我们可以把它递归地分割成两个非空子字符串,从而将其表示为二叉树. 下图是字符串 s1 = "great" 的一种可能的表示形式. great / \ gr eat / \ / \ g r e at / \ a t 在扰乱这个字符串的过程中,我们可以挑选任何一个非叶节点,然后交换它的两个子节点. 例如,如果我们挑选非叶节点 "gr" ,交换它的两个子节点,将会产生扰乱字符串 "rgeat" . rgeat…
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…
1.2 Implement a function void reverse(char *str) in C or C++ which reverses a null-terminated string. 这道题让我们用C++或C语言来翻转一个字符串,不算一道难题,在之前那道Reverse Words in a String 翻转字符串中的单词中用到了这个函数,跟那道题比起来,这题算简单的了.C语言的版本要比C++的稍微复杂一些,应为string类集成了很多有用的功能,比如得到字符串的长度,用下标…
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…
help 命令,3种形式: help 命令 形式 help @<group> 比如:help @generic.help @string.help @hash.help @list.help @set.help @sorted_set.help @transactions 等 help <command> help <tab> help @group 命令中,group 的类型: group 类型 备注 cluster help @cluster,查看集群信息 gene…