[Leetcode] Scramble String

最新推荐文章于 2019-08-26 13:50:26 发布

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本文探讨了如何使用动态规划（DP）和递归来解决字符串乱序问题，对比了两种方法的效率，并提供了代码实现。通过实例分析，展示了如何判断两个字符串是否为彼此的乱序字符串。

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.

目前 Leetcode 上最难的 DP 题，三维 DP。

answer[i][j][k] 表示 S1.substr(i,k) 与 S2.substr(j,k) 是否互为 scramble string。

DP 的状态方程与代码参考：http://blog.youkuaiyun.com/sunjilong/article/details/8266313

然而对这道题而言，用递归远比 DP 来得直观；普通的递归会大集合超时，然而在加入剪枝后，performance 似乎也与 DP 差不离了。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        assert(s1.length()==s2.length());
        //base case
        if(s1==s2) return true;    
        
        //剪枝
        string sort1 = s1, sort2 = s2;
        sort(sort1.begin(),sort1.end());
        sort(sort2.begin(),sort2.end());
        if(sort1!=sort2) return false;
        
        for(int i=1;i<s1.length();i++)
        {
            string subStr11 = s1.substr(0,i);
            string subStr12 = s1.substr(i);
            
            string subStr21 = s2.substr(0,i);
            string subStr22 = s2.substr(i);
            
            if(isScramble(subStr11,subStr21)&&isScramble(subStr12,subStr22))
            return true;
            
            subStr21 = s2.substr(s2.length()-i);
            subStr22 = s2.substr(0,s2.length()-i);
            
            if(isScramble(subStr11,subStr21)&&isScramble(subStr12,subStr22))
            return true;
        }       
        
        return false;
    }
};