[LeetCode] Scramble String

最新推荐文章于 2019-08-26 13:50:26 发布

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本文探讨了如何判断两个字符串是否可以通过乱序匹配成为彼此，详细介绍了算法实现过程及核心逻辑。

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.

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size() != s2.size()) return false;
        if(s1 == s2) return true;
        int A[26] = {0};
        for(int i = 0;i < s1.size();i ++)
            A[s1[i] - 'a'] ++;
        for(int i = 0;i < s2.size();i ++)
            A[s2[i] - 'a'] --;
        for(int i = 0;i < 26;i ++)
            if(A[i] != 0)
                return false;
        for(int i = 1;i < s1.size();i ++){
            bool result = isScramble(s1.substr(0,i),s2.substr(0,i)) && isScramble(s1.substr(i),s2.substr(i));
            result = result || isScramble(s1.substr(0,i),s2.substr(s2.size() - i,i)) && isScramble(s1.substr(i),s2.substr(0,s2.size() - i));
            if(result) return true;
        }
        return false;
    }
};