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 choose any non-leaf node and swap its two children.
For example, if we choose the node "gr"
and swap its two children, it produces a scrambled
string "rgeat"
.
rgeat / rg eat / \ / r g e at / a t
We say that "rgeat"
is a scrambled string of "great"
.
Similarly, if we continue to swap the children of nodes "eat"
and "at"
,
it produces a scrambled string "rgtae"
.
rgtae / rg tae / \ / r g ta e / t a
We say that "rgtae"
is a scrambled string of "great"
.
Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.
This problem can be solved with brute Force.
class Solution: # @param s1, a string # @param s2, a string # @return a boolean def isScramble(self, s1, s2): n, m = len(s1),len(s2) if n!=m or sorted(s1)!=sorted(s2): return False if n < 4 or s1 == s2: return True for i in range(1, n): if (self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:], s2[i:]) )or (self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:], s2[:-i]) ): return True return False
时间: 2024-11-05 18:53:16