Scramble String

最新推荐文章于 2021-04-30 19:30:28 发布

最新推荐文章于 2021-04-30 19:30:28 发布 · 84 阅读

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本文探讨了如何通过递归划分字符串并交换子串位置来判断一个字符串是否为另一个字符串的乱序排列，深入理解字符串操作与算法逻辑。

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.

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if (s1.length() != s2.length()) {
        	return false; 
        }
        if (s1.length()==1 && s2.length()==1) {
        	return s1.charAt(0) == s2.charAt(0); 
        }
        char[] arr1 = s1.toCharArray();
        char[] arr2 = s2.toCharArray();
        Arrays.sort(arr1);
        Arrays.sort(arr2);
        if (!new String(arr1).equals(new String(arr2))) {
        	return false;
        }
        if (s1.equals(s2)) {
        	return true;
        }
        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, s2.length() - split);
            s22 = s2.substring(s2.length() - split);
            if(isScramble(s11, s22) && isScramble(s12, s21)) {
            	return true;
            }
        }
        return false;
    }
}