LeetCode开心刷题三十三天——87. Scramble String

小鱼儿 2023-08-17 15:14 99阅读 0赞

87. Scramble String

Hard

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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
    
    ATTENTION:
    when we recursively call the function itself.use self before it.
    use range remember it has two parameters,begin&end
    During solving this problem I face the error report cannot find the 
    function because I didn't add self before the function I recursive.On the other
    hand when you face a function or variable error ,find all possible places.
    Such as me,this time only find one,so don't discover problem immediately

class Solution(object):
        def isScramble(self, s1, s2):
            N = len(s1)
            if N == 0:
                return True
            if N == 1:
                return s1 == s2
            if sorted(s1) != sorted(s2):
                return False
            for i in range(1,N):
                if self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:],s2[i:]):
                    return True
                elif self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:],s2[:-i]):
                    return True
            return False
    
    solu = Solution()
    s1 = "great"
    s2 = "rgeat"
    print(solu.isScramble(s1,s2))

Detailed Comment:

详细注释版

class Solution(object):
        def isScramble(self, s1, s2):
            """
            :type s1: str
            :type s2: str
            :rtype: bool
            """
            # NULL
            n=len(s1)
            # easy forget Judge NULL
            if sorted(s1)!=sorted(s2):
                return False
            if n==0:
                return 1
            if n==1:
                return s1==s2
            # range from 1 easy miss s[:0]bring error
            # actual meaning i-->partition string.0 partition no meaning
            for i in range(1,n):
                # core-->length==i
                # structure directly return only run once.just judge finally return
                if self.isScramble(s1[:i],s2[:i]) and self.isScramble(s1[i:],s2[i:]):
                    return True
                # i range need combine to full.array range length before&after and needn't equal,equal meaningless
                elif self.isScramble(s1[:i], s2[-i:]) and self.isScramble(s1[i:], s2[:-i]):
                    return True
            return False

转载于:https://www.cnblogs.com/Marigolci/p/11338093.html