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87. Scramble String

难度:Hard

刷题内容

原题连接

  • https://leetcode.com/problems/scramble-string/

内容描述

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.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true
Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

˼·1 **- ʱ¼ä¸´ÔÓ¶È: O(2^n)*- ¿Õ¼ä¸´ÔÓ¶È: O(1)***

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

˼·2 **- ʱ¼ä¸´ÔÓ¶È: O(n^4)****- ¿Õ¼ä¸´ÔÓ¶È: O(n^3

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class Solution {
public:
    bool isScramble(string s1, string s2) {
    int len = s1.length();
    int dp[len][len][len + 1] = {0};
    for(int i = len - 1;i >= 0;--i)
        for(int j = len - 1;j >= 0;--j)
            for(int t = 1;t <= min(len - i,len - j);++t)
                if(t == 1)
                    dp[i][j][t] = (s1[i] == s2[j]);
                else
                {
                    int k = 1;
                    for(;k < t;++k)
                        if((dp[i][j][k] && dp[i + k][j + k][t - k]) || (dp[i][j + t - k][k] && dp[i + k][j][t - k]))
                            break;
                    dp[i][j][t] = k == t ? 0 : 1;
                }
    return dp[0][0][len];

}
};
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