描述
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.
分析
这道题我已经弃疗了,三维动态规划,正确解摆在面前也费劲。参考这里。
有一个递归的解写起来比较容易,但是会超时。
代码
动态规划
1 |
class (object): |
递归(超时)
1 |
class (object): |
近期评论