LeetCode| Scramble String

最新推荐文章于 2019-06-05 11:26:04 发布

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本文介绍了一种判断两个等长字符串是否可通过特定规则相互转换的算法。通过递归与动态规划相结合的方式，解决字符串的乱序匹配问题。该算法考虑了字符串作为二叉树结构时节点的交换操作。

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.

思路：

用递归可以非常简单地完成，但是递归时需要比较两种情况：

1）s1左侧与s2左侧比较，s1右侧与s2右侧比较

2）s1左侧与s2右侧比较，s1右侧与s2左侧比较

为了简化计算量，我们可以用动态规划的方法来完成。dp[l][i][j]中的l+1代表子字符串的长度，i代表s1的起始点，j代表s2的起始点。

代码：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        
        if(s1.size()!=s2.size())
        {
            return false;
        }
        else
        {
            int len = s1.size();
            bool*** dp = new bool**[len];
            for(int i=0;i<len;i++)
            {
                dp[i]=new bool*[len];
                for(int j=0;j<len;j++)
                {
                    dp[i][j]=new bool[len];
                }
            }
            
            for(int i=0;i<len;i++)
            {
                for(int j=0;j<len;j++)
                {
                    dp[0][i][j]=(s1[i]==s2[j]);
                }
            }
            
            for(int l=2;l<=len;l++)
            {
                for(int i=0;i+l<=len;i++)
                {
                    for(int j=0;j+l<=len;j++)
                    {
                        dp[l-1][i][j] = false;
                        for(int k=1;k<=l-1;k++)
                        {
                            if(dp[k-1][i][j]&&dp[l-k-1][i+k][j+k])
                            {
                                dp[l-1][i][j]=true;
                            }
                            else if(dp[k-1][i][j+l-k]&&dp[l-k-1][i+k][j])
                            {
                                dp[l-1][i][j]=true;
                            }
                        }
                    }
                }
            }
            
            return dp[len-1][0][0];
        }
    }
};