leetcode : Scramble String

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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.

很明显要用动态规划解，注意s1跟s2的长度一定是相等的。

问题类似于钢条切割；只不过把钢条最大收益变成了是否是scrambled string

要记录s1中的某一段与s2中的某一段是否满足scrambled string条件，需要分别记录两段的起点于中点，但是，如果个子串满足条件，那么他们的长度必然是一样的。

所以备忘录的结构为三维数组，即子串1的起点i，子串2的起点j，以及串长k。即bool dp[i][j][k]

dp[i][j][k]  =  (dp[i][j][x] && dp[i + x][j + x][k - x]) || (dp[i][j + k - x][x] && dp[i + x][j][k - x])

k = 1的情况即边界情况

代码：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int length = s1.length();  
        bool f[length][length][length];  
        memset(f, false, sizeof(bool) * length * length * length);  
          
        for (int k = 1; k <= length; k++) {  
            for (int i = 0; i <= length - k; i++) {  
                for (int j = 0; j <= length - k; j++) {  
                    if (k == 1) {  
                        f[i][j][k] = s1[i] == s2[j];  
                    }  
                    else {  
                        for (int l = 1; l < k; l++) {  
                            if ((f[i][j][l] && f[i + l][j + l][k - l]) || (f[i][j + k - l][l] && f[i + l][j][k - l])) {  
                                f[i][j][k] = true;  
                                break;  
                            }                              
                        }  
                    }  
                }  
            }              
        }  
                  
        return f[0][0][length];  
    }  
};

leetcode : Scramble String

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原文地址：http://www.cnblogs.com/64open/p/4122031.html