[LeetCode] 87. Scramble String_hard tag: 区间Dynamic Programming?

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