leetcode 87. Scramble String

最新推荐文章于 2023-03-30 18:15:00 发布

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最新推荐文章于 2023-03-30 18:15:00 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。通过递归分割字符串并验证子串是否能通过重新排列及内部节点子串交换达到目标字符串的状态，实现了对Scramble String问题的有效求解。

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87. Scramble String

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.

有一点递归的感觉。

class Solution {
public:
    bool iscontain(string s1,string s2)
    {
        sort(s1.begin(), s1.end());
        sort(s2.begin(), s2.end());
        return (s1 == s2);
    }
    bool ischeck(string s1,string s2)//第i个的前面分割，s2作为target
    {
        if (s1 == s2)
            return 1;
        if ( !iscontain(s1, s2))      
            return 0;
        int tsize = s2.size();
        for (int i = 1; i < tsize; i++)    
        {    
            if (ischeck(s1.substr(0,i), s2.substr(0,i)) && ischeck(s1.substr(i, tsize-i),s2.substr(i, tsize-i)))
                return 1;
            if (ischeck(s1.substr(0,i), s2.substr(tsize-i, i)) && ischeck(s1.substr(i, tsize-i),s2.substr(0, tsize-i)))//交换
                return 1;
        } 
        return 0;
    }
    
    bool isScramble(string s1, string s2) //要有一个target
    {
        if (s1.size() != s2.size())
            return 0;
        return ischeck(s1, s2);
    }
};