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.
字符串的好题。题干解释的非常复杂,一下让人不知所措了。
这道题到底是什么意思呢?最终的结果是把一个字符串中字母的顺序打乱了,让你判断一个字符串能不能由另一个字符串打乱得到。那打乱这个过程是怎么做的呢,很简单,给你一个字符串,你必须先找一个点把它砍成两半,你可以通过交换这两半的顺序来打乱源字符串的顺序,也就是在两半中的字符与另一半中所有字符的相对顺序是统一的。对于每一半,都可以重复上面的过程。
那想一下,怎么知道打断的那个点在哪呢?穷举。怎么知道打断之后有没有做交换操作呢?两种情况递归,有一条走的通就可以了。还有个问题,两个字符串中包含的字符一定是完全一样的,怎样确定这一点呢?最暴力的方式,新开两个字符串,排序,判断这两个新的相不相等。
比如: "abcd", "bdac" are not scramble string
代码:
class Solution: # @return a boolean def isScramble(self, s1, s2): if s1==s2: return True if sorted(s1) != sorted(s2): return False # "abcd", "bdac" are not scramble string length=len(s1) for i in range(1,length): if self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:]): return True if self.isScramble(s1[:i],s2[length-i:]) and self.isScramble(s1[i:],s2[:length-i]): return True return False