【ATT】Scramble String

最新推荐文章于 2025-09-10 15:24:29 发布

转载最新推荐文章于 2025-09-10 15:24:29 发布 · 52 阅读

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原文链接：http://www.cnblogs.com/summer-zhou/p/3337814.html

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#数据结构与算法 #c/c++

本文介绍了一种解决字符串Scramble问题的方法，通过三维动态规划来判断两个字符串是否可以通过交换子串的方式相互转换。文章详细阐述了动态规划的状态定义、转移方程及实现细节。

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.

A: 三维动态规划。

最初的想法是用递归去做，发现有很多重叠的子问题。因此改用动态规划去存储中间状态。

dp[i][j][k]：初始位置为i，长度为k的s1子串和初始位置为j，长度为k的s2子串是否scramble的。

边界条件是dp[i][j][1].

k>=2时，

计算dp[i][j][k]时，对于任意的l<k,>0,满足子串s1[i~,i+l-1]和s2[j,j+l-1]scramble，同时剩余的两部分也scramble；或者s1[i+k-l,i+k-1]和s2[j,j+l-1]scramble,同时剩余的两部分也scramble。！！注意，不要只考虑一种情况。

复杂度O(n^4).

    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s1.empty()&&s2.empty())
            return true;
            
        int n = s1.size();
        
        vector<vector<vector<bool>>> dp(n,vector<vector<bool>> (n,vector<bool>(n+1,false)) );
        
        int i,j,k,l;
        for(i=0;i<n;i++)
            for(j=0;j<n;j++)
                dp[i][j][1] = s1[i]==s2[j];
                
        for(k=2;k<=n;k++)
        {
            for(i=0;i<=n-k;i++)
            {
                for(j=0;j<=n-k;j++)
                {
                    for(l=1;l<k;l++)
                    {
                        dp[i][j][k] = (dp[i][j][l]&&dp[i+l][j+l][k-l])||(dp[i][j+k-l][l]&&dp[i+l][j][k-l]);
                        if(dp[i][j][k])
                            break;
                    }
                }
            }
        }
        
        return dp[0][0][n];
        
    }

转载于:https://www.cnblogs.com/summer-zhou/p/3337814.html