Scramble String

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本文介绍了一种用于判断两个字符串是否互为ScrambleString的算法。通过构建二叉树来表示字符串，并使用动态规划的方法，递归地检查所有可能的子字符串组合。文章详细解释了动态规划的状态定义、初始化及状态转移方程。

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”:
这里写图片描述
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“.

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“.
这里写图片描述

解题思想：采用动态规划的思想,假设两字符串是s1,s2,判断s1是否是s2的scrambled string。
dp[i][j][len]:从s1[i]和s2[j]开始长度为len的子字符串是否是scrambled的。
初始条件：dp[i][j][1]=(s1[i]==s2[j]);
递推公式：k=1:len-1
dp[i][j][len]|=(dp[i][j][k]&&dp[i][j][len-k])
||(dp[i][j+len-k][k]&&dp[i+k][j][len-k]); //判断前后交换的情况

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size()!=s2.size())
            return false;
        if(s1==s2)
            return true;
        int n=s1.size();
        bool dp[n][n][n+1];
        memset(dp,0,sizeof(dp));
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                dp[i][j][1]=(s1[i]==s2[j]);

        for(int len=2;len<=n;len++)
            for(int i=0;i<n-len+1;i++)
                for(int j=0;j<n-len+1;j++)
                    for(int k=1;k<len;k++)
                        dp[i][j][len]|=(dp[i][j][k]&&dp[i+k][j+k][len-k])||    
                        (dp[i][j+len-k][k]&&dp[i+k][j][len-k]);

        return dp[0][0][n];
    }
};