Scramble String

最新推荐文章于 2019-06-05 11:26:04 发布

jialuyyy

最新推荐文章于 2019-06-05 11:26:04 发布

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分类专栏： recursion

本文链接：https://blog.youkuaiyun.com/jialuyyy/article/details/46754257

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本文深入探讨了如何通过递归解决ScrambleString问题，即判断两个字符串是否可以通过交换子节点来形成乱序匹配。具体步骤包括预处理、长度检查、字符频率比较以及递归验证子字符串的匹配。此外，文章还强调了解决此类问题时需考虑的特殊情况，如字符串长度不等、字符不同或完全相同的情况。

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Scramble String

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.

Solution

It seems that this problem is not easy to solve. However, if we use recursion to solve this problem, it is not difficult at all.

If string1 is a scrambled string of string2, we can break the two strings into two partition. string1 becomes string11 and string12 while string2 becomes string21 and string22.

If scramble(string11, string21) && scramble(string12, string22) is true, or scramble(string11, string22) && scramble(string21, string12) is true, then string1 is the scrambled string of string2.

For example,

"great" and "rgeat"

great -> "gr" "eat"

rgeat -> "rg" "eat" (string11, string21) && (string12, string22) == true

For another example,

"rgtae" and "great"

For the substring "tae" and "eat",

"tae" -> "ta" and "e"

"eat" -> "e" and "at" (string11, string22) &&(string12, string21) == true

If we do not do any preprocessing for this problem, the time will be out, so we need to consider some cases:

1. If the length of two strings are not equal, false;

2. If two strings has different characters, false;

3. If two strings are totally the same, true.

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.length() != s2.length())
            return false;
        
        if (s1.equals(s2))
            return true;
        
        int[] ch = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            ch[s1.charAt(i) - 'a']++;
            ch[s2.charAt(i) - 'a']--;
        }
        
        for (int i = 0; i < 26; i++) {
            if (ch[i] != 0)
                return false;
        }
        
        for (int split = 1; split < s1.length(); split++) {
            String s11 = s1.substring(0, split);
            String s12 = s1.substring(split);
            
            String s21 = s2.substring(0, split);
            String s22 = s2.substring(split);
            
            if (isScramble(s11,s21) && isScramble(s12,s22)) return true;
            
            s21 = s2.substring(0, s1.length() - split);
            s22 = s2.substring(s1.length() - split);
            
            if (isScramble(s11,s22) && isScramble(s12,s21)) return true;
        }
        return false;
    }
}