天题系列： Scramble String -- 三维动态规划

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.

 

参考：

http://www.blogjava.net/sandy/archive/2013/05/22/399605.html 引用一下题解部分

http://blog.csdn.net/fightforyourdream/article/details/17707187

“对付复杂问题的方法是从简单的特例来思考，从而找出规律。
先考察简单情况：
字符串长度为1：很明显，两个字符串必须完全相同才可以。
字符串长度为2：当s1="ab", s2只有"ab"或者"ba"才可以。
对于任意长度的字符串，我们可以把字符串s1分为a1,b1两个部分，s2分为a2,b2两个部分，满足（(a1~a2) && (b1~b2)）或者 （(a1~b2) && (a1~b2)）”

 

“
这里我使用了一个三维数组boolean result[len][len][len],其中第一维为子串的长度，第二维为s1的起始索引，第三维为s2的起始索引。result[k][i][j]表示s1[i...i+k]是否可以由s2[j...j+k]变化得来”

        if(s1==null || s2==null||s1.length()!=s2.length()) return false;

        int len = s1.length();

        boolean[][][] dp = new boolean[len][len][len];

        char[] c1 = s1.toCharArray();

        char[] c2 = s2.toCharArray();

        

        for(int i=0;i<len;i++){

            for(int j=0;j<len;j++){

                dp[0][i][j] = c1[i]==c2[j];

            }

        }

        for(int k=2;k<=len;k++){ 

            for(int i=len-k;i>=0;i--){ // the order is from the base, i.e. begin from the last position

                for(int j=len-k;j>=0;j--){

                    boolean r = false;  // test if cut anywhere within k length, the scramble exists

                    for(int cut=1;!r&&cut<k;cut++){ // !r&& 是因为必须每种cut都通过

                        r = (dp[cut-1][i][j]&&dp[k-cut-1][i+cut][j+cut])||(dp[cut-1][i][j+k-cut]&&dp[k-cut-1][i+cut][j]); //前前 && 后后 || 前后 && 后前

                    }

                    dp[k-1][i][j]=r; // can r go through all cuts within k?

                }

            }

        }

        return dp[len-1][0][0];

    }

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