【LeetCode】Scramble String

最新推荐文章于 2019-08-26 13:50:26 发布

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本文介绍了一种用于判断两个字符串是否可以通过乱序操作相互转换的算法。通过递归或动态规划的方法，对字符串进行分割并比较分割后的子串，利用哈希表优化性能，避免了全排列的暴力搜索。

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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

题解：递归或动态规划。首先用暴力搜索整个解答树，刚开始往二叉树那方面想越想越复杂，实际上只需要转化为子问题，也就是给定一个串如何递归成子方法来解决，用遍历分割思想，对s1遍历用i来分成两部分，注意这里有两种情况一种是第一部分对应相同另一种是交换对应相同，这里关键是用剪枝来剪掉大量非法串——最简单的思想就是这两个串“不一样”,这里不一样是指每个字母出现次数，因此可以用hash记录次数，对s1出现字母++，s2出现字母--从而只需判断是否为0，也可以排序但是时间开销大。至于dp需要用到三维，dp[i][j][len]i是s1头指针，j是s2指针，len是选取长度

代码：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1==s2)
            return true;
        if(s1.size()!=s2.size()) return false;
        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;
    }
};