LeetCode Scramble String

字符串乱序匹配算法

最新推荐文章于 2019-08-26 13:50:26 发布

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本文介绍了一种判断两个等长字符串是否可通过特定规则相互转换的算法。通过递归划分字符串并利用动态规划方法，实现对乱序字符串的有效匹配。文章提供了完整的C++代码实现。

题目：

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.

分析及代码参考： http://cozilla.iteye.com/blog/1950242

class Solution {
public:
	//DP,dp[i][j][k]表示从s1[i]开始长度为k的字符串与从s2[j]开始长度为k的字符串是否满足scramble
	//dp[i][j][k] = (dp[i][j][t]&&dp[i+t][j+t][k-t]) || (dp[i][j+k-t][t]&&dp[i+t][j][k-t])，其中1<=t<k;
	bool isScramble(string s1, string s2) {
		int n = s1.size();
		bool dp[100][100][100] = { false };
		//初始化
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				dp[i][j][0] = true;
				dp[i][j][1] = (s1[i] == s2[j]);
			}
		}
		for (int i = n-1; i >=0; i--) {
			for (int j = n-1; j >=0; j--) {
				for (int k = 2; (i + k <= n) && (j + k <= n); k++) {
					for (int t = 1; t < k; t++) {
						if (dp[i][j][k])
							break;
						else
							dp[i][j][k] = (dp[i][j][t] && dp[i + t][j + t][k - t]) || (dp[i][j + k - t][t] && dp[i + t][j][k - t]);
					}
				}
			}
		}
		return dp[0][0][n];
	}
};