87. Scramble String

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最新推荐文章于 2023-03-30 18:15:00 发布

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分类专栏： leetcode 文章标签： string

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本文介绍了一种算法，用于判断一个字符串是否可以通过特定的搅乱规则从另一个字符串转换而来。通过递归地划分字符串并检查子串是否符合搅乱规则来实现。

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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.

给出两个字符串，判断第二个字符串是否能从第一个字符串“搅乱“得来。“搅乱“的规则是对每个字符串递归的进行旋转（不旋转也行）两个子字符串。用递归的方法即可。通过递归划分子问题：对当前两个字符串s1，s2，如果s1(0,i)和s2(0,i)符合“搅乱”规则且s1(i,n1)和s2(i,n2)符合“搅乱”规则（交换），或者s1(0,i)和s2(n2-i,n2)符合“搅乱”规则且s1(i,n1)和s2(0,i)符合“搅乱”规则,则说明当前两个字符串符合“搅乱”规则。在递归过程中为了减少递归次数，通过检测两个字符串字符数和所含字符是否全都一样来剪枝。

代码：

class Solution
{
public:
	bool isScramble(string s1, string s2)
	{
		if(s1 == s2) return true;
		int n1 = s1.size(), n2 = s2.size();
		if(n1 != n2) return false;

		int cnt[256] = {0};
		for(int i = 0; i < n1; ++i)
		{
			++cnt[s1[i]];
			--cnt[s2[i]];
		}
		for(int i = 0; i < 256; ++i)
		{
			if(cnt[i] != 0)
			{
				return false;
			}
		}

		for(int i = 1; i < n1; ++i)
		{
			if((isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i))) 
				|| (isScramble(s1.substr(0, i), s2.substr(n2 - i)) && isScramble(s1.substr(i), s2.substr(0, n2 - i))) )
			{
				return true;
			}
		}
		return false;
	}
};