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.
Example 1:
Input: s1 = "great", s2 = "rgeat" Output: true
Example 2:
Input: s1 = "abcde", s2 = "caebd" Output: false
这个题目我们先判断s1, s2的长度是否相等,然后s1 == s2 否,接着如果s1和s2的所含字符不一样,也肯定不是,接着进入正式的判断搜索,然后分两种情况,s2的后半段有无scramble
1) 无scramble i = [1, n]
isScramble(s1[:i], s2[:i]) and isScramble(s1[i:], s2[i:])
2) 有scramble
isScramble(s1[:i], s2[-i: ]) and isScramble(s1[i:], s2[: -i])
Code
class Solution: def isScramble(self, s1, s2): m, n = len(s1), len(s2) if m != n: return False if s1 == s2: return True if sorted(list(s1)) != sorted(list(s2)): return False 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