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…

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 cho…

(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…

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 an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is guaranteed to be a positive integer. You may assume th…

A magical string S consists of only '1' and '2' and obeys the following rules: The string S is magical because concatenating the number of contiguous occurrences of characters '1' and '2' generates the string S itself. The first few elements of strin…