LeetCode 0079 - Word Search

Reck Zhang

发布于 2021-08-11 10:51:02

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发布于 2021-08-11 10:51:02

文章被收录于专栏：Reck Zhang

Scramble String

Desicription

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.

Solution

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1 == s2)
            return 1;
        vector<int> cnt(26, 0);
        for(int i = 0; s1[i]; i++)
            cnt[s1[i] - 'a']++, cnt[s2[i] - 'a']--;
        for(int i = 0; i < 26; i++)
            if(cnt[i])
                return 0;
        for(int i = 1; s1[i]; i++){
            if(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
                return 1;
            if(isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i), s2.substr(0, s1.size() - i)))
                return 1;
        }
        return 0;
    }
};

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原始发表：2017-11-21，如有侵权请联系 cloudcommunity@tencent.com 删除

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