leetcode 87. Scramble String 字符串拼凑 && DFS深度优先搜索

冷不防 2022-06-08 02:29 143阅读 0赞

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.

这道题题意很简单，但是我没有想到深度优先搜索可以这么做，我是网上参考的答案，这个解决方法很漂亮。

建议和[leetcode 95. Unique Binary Search Trees II 递归构造所有可能的搜索二叉树BST + 卡特兰数][leetcode 95. Unique Binary Search Trees II _BST _] 一起学习，因为感觉递归思路相识

这道题的做法感觉和[leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal 中前序构造BST][leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal _BST] 和 [leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal 中后序构造BST][leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal _BST] 一起学习，因为做法类似

代码如下：

public class Solution 
    {
        /*
         * 暴力求解，这个我就真的想不到了
         * */
        public boolean isScramble(String s1, String s2)
        {
            if(s1==s2 || s1.equals(s2))
                return true;
    
            int []a=new int[26];
            for(int i=0;i<s1.length();i++)
                a[s1.charAt(i)-'a']++;
            for(int i=0;i<s2.length();i++)
                a[s2.charAt(i)-'a']--;
    
            for(int i=0;i<a.length;i++)
            {
                if(a[i]!=0) 
                    return false;
            }
    
            //注意这里是  <
            for(int i=1;i<s1.length();i++)
            {
                if(isScramble(s1.substring(0,i), s2.substring(0,i)) 
                && isScramble(s1.substring(i), s2.substring(i)))
                    return true;
    
                if(isScramble(s1.substring(0,i), s2.substring(s2.length()-i)) 
                && isScramble(s1.substring(i), s2.substring(0,s2.length()-i)))
                    return true;
            }
            return false;
        }
    }

下面是C++的做法，就是使用DFS深度优先遍历暴力搜索，说实话，我自己想不出来，这是网上看的做法，很棒

代码如下:

#include <iostream>
    #include <string>
    #include <vector>
    
    using namespace std;
    
    class Solution 
    {
    public:
        bool isScramble(string s1, string s2) 
        {
            if (s1 == s2)
                return true;
            if (s1.length() != s2.length())
                return false;
    
            vector<int> count(26,0);
            for (int i = 0; i < s1.length(); i++)
                count[(int)(s1[i] - 'a')]++;
            for (int i = 0; i < s2.length(); i++)
                count[(int)(s2[i] - 'a')]--;
            for (int i = 0; i < count.size(); i++)
            {
                if (count[i] != 0)
                    return false;
            }
    
            for (int i = 1; i < s1.length(); 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(s2.length()-i))
                    && isScramble(s1.substr(i), s2.substr(0, s2.length()-i)))
                    return true;
            }
            return false;
        }
    };

[leetcode 95. Unique Binary Search Trees II _BST _]: http://blog.csdn.net/jackzhang_123/article/details/77948273
[leetcode 105. Construct Binary Tree from Preorder and Inorder Traversal _BST]: http://blog.csdn.net/jackzhang_123/article/details/77977697
[leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal _BST]: http://blog.csdn.net/jackzhang_123/article/details/77977904