[leetcode] Scramble String

最新推荐文章于 2019-08-26 13:50:26 发布

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本文介绍了一种用于判断字符串是否为另一个字符串乱序排列版本的算法，通过递归和动态规划方法解决字符串匹配问题。

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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.

递归的做法：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        return _Scramble(s1,s2);
    }
    
    bool _Scramble(string s1, string s2){
        int length1=s1.size();
        int length2=s2.size();
        if(length1!=length2)
            return false;
        string t1=s1;
        string t2=s2;
        sort(t1.begin(),t1.end());
        sort(t2.begin(),t2.end());
        if(t1!=t2)
            return false;
        if(length1==1 && t1==t2)
            return true;
        if(length1==2 && t1==t2)
            return true;
        for(int i=1 ; i<length1 ; i++){
            string s11(s1.begin(),s1.begin()+i);
            string s12(s1.begin()+i,s1.end());
            string s21(s2.begin(),s2.begin()+i);
            string s22(s2.begin()+i,s2.end());
            string s31(s2.begin(),s2.begin()+length1-i);
            string s32(s2.begin()+length1-i,s2.end());
            if(_Scramble(s11,s21)&&_Scramble(s12,s22))
                return true;
            if(_Scramble(s11,s32)&&_Scramble(s12,s31))
                return true;
        }
        return false;
    }
};

可以用动态规划去做，敬请期待