LeetCode 87. Scramble String（字符串扰乱）

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本文介绍了一种用于判断两个字符串是否为乱序版本的方法。该方法采用递归策略，并利用字符频率直方图进行优化，以提高效率。具体实现包括了如何划分字符串、交换子串并验证乱序字符串的有效性。

原题网址：https://leetcode.com/problems/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.

方法：递归。注意通过频率统计直方图进行剪枝。

public class Solution {
    private int[] frequency(char[] sa, int from, int to) {
        int[] f = new int[26];
        for(int i=from; i<=to; i++) {
            f[sa[i]-'a'] ++;
        }
        return f;
    }
    private boolean equalFrequency(int[] f1, int[] f2) {
        for(int i=0; i<f1.length; i++) {
            if (f1[i] != f2[i]) return false;
        }
        return true;
    }
    private boolean scramble(char[] sa1, int f1, int t1, char[] sa2, int f2, int t2) {
        if (t1 - f1 != t2 - f2) return false;
        if (f1>t1) return false;
        if (f1==t1) return sa1[f1] == sa2[f2];
        int[] c1 = frequency(sa1, f1, t1);
        int[] c2 = frequency(sa2, f2, t2);
        if (!equalFrequency(c1, c2)) return false;
        for(int i=0; i<t1-f1; i++) {
            if (scramble(sa1, f1, f1+i, sa2, f2, f2+i) && scramble(sa1, f1+i+1, t1, sa2, f2+i+1, t2)) return true;
            if (scramble(sa1, f1, f1+i, sa2, t2-i, t2) && scramble(sa1, f1+i+1, t1, sa2, f2, t2-i-1)) return true;
        }
        return false;
    }
    public boolean isScramble(String s1, String s2) {
        char[] sa1 = s1.toCharArray();
        char[] sa2 = s2.toCharArray();
        return scramble(sa1, 0, sa1.length-1, sa2, 0, sa2.length-1);
    }
}