Scramble String

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题目

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

该题目用到了三维动态规划，到目前为止没有理解深入。

参考了：http://blog.csdn.net/jiyanfeng1/article/details/8620224

http://www.cnblogs.com/remlostime/archive/2012/11/19/2778108.html

    public boolean isScramble(String s1, String s2) {    if (s1.equals(s2)) {    return true;    }    int len1 = s1.length();    int len2 = s2.length();    boolean[][][] scrambled = new boolean[len1][len2][len1 + 1];    for (int i = 0; i < len1; i++) {    for (int j = 0; j < len2; j++) {    scrambled[i][j][0] = true;    scrambled[i][j][1] = (s1.charAt(i) == s2.charAt(j));    }    }        for (int i = len1 - 1; i >= 0; i--) {    for (int j = len2 - 1; j >= 0; j--) {    for (int n = 2; n <= Math.min(len1 - i, len2 - j ); n++) {    for (int m = 1; m < n; m++) {                        scrambled[i][j][n] |= scrambled[i][j][m] && scrambled[i + m][j + m][n - m] ||                                  scrambled[i][j + n - m][m] && scrambled[i + m][j][n - m];                          if(scrambled[i][j][n])  break;      }    }    }    }    return scrambled[0][0][len1];    }

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