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87. scramble string

admin 11月 11, 2020 0

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”:

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    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”.

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    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”.

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    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.

解法1： 递归　O(2^n)

尝试每一个可能的split point，然后查看对于某一个切口i,可能可以分成的４个substring是否是scramble string.
时间复杂度是O(2^N)

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class  {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) {
            return true;
        }
        if (s1.length() != s2.length()) {
            return false;
        }
        int[] chars = new int[256];
        for (int i = 0; i < s1.length(); i++) {
            chars[(int)s1.charAt(i)]++;
            chars[(int)s2.charAt(i)]--;
        }
        for (int i = 0; i < 256; i++) {
            if (chars[i] != 0) return false;
        }
        for (int i = 1; i < s1.length(); i++) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i))
                && isScramble(s1.substring(i), s2.substring(i)))
                return true;
            if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i))
               && isScramble(s1.substring(i), s2.substring(0, s2.length() - i)))
                return true;
        }
        return false;
    }
}

解法２: DP

参考了[这篇][1]帖子的解法，　可以用dp的思想，dp是一个bottom up的想法。
用一个dp[i][j][k]表示两个string分别从i和j开始，长度为k + 1的两个子串是否为scramble string

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public boolean isScramble(String s1, String s2) {  
  int len = s1.length();  
  if (len != s2.length()) return false;  
  if (s1.equals(s2)) return true;  
  
  // T[i][j][k] = true iff s2.substring(j,j+k+1) is a scambled string of s1.substring(i,i+k+1)  
  boolean[][][] scrambled = new boolean[len][len][len];  
  for (int i=0; i < len; ++i) {  
    for (int j=0; j < len; ++j) {  
      scrambled[i][j][0] = (s1.charAt(i) == s2.charAt(j));  
    }  
  }  
  // dynamically fill up the table  
  for (int k=1; k < len; ++k) { // k: length  
    for (int i=0; i < len - k; ++i) { // i: index in s1  
      for (int j=0; j < len - k; ++j) { // j: index in s2  
        scrambled[i][j][k] = false;  
        for (int p=0; p < k; ++p) { // p: split into [0..p] and [p+1..k]  
          if ((scrambled[i][j][p] && scrambled[i+p+1][j+p+1][k-p-1])  
              || (scrambled[i][j+k-p][p] && scrambled[i+p+1][j][k-p-1])) {  
            scrambled[i][j][k] = true;  
            break;  
          }  
        }  
      }  
    }  
  }  
  return scrambled[0][0][len-1];  
}
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