LeetCode刷提笔录Scramble String

最新推荐文章于 2018-12-11 09:53:55 发布

原创最新推荐文章于 2018-12-11 09:53:55 发布 · 897 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #动态规划

LeetCode 同时被 2 个专栏收录

114 篇文章

订阅专栏

动态规划

14 篇文章

订阅专栏

本文深入探讨了如何使用三维动态规划（DP）算法解决字符串乱序匹配问题，详细解释了算法背后的逻辑、核心步骤以及实现过程。通过实例分析，读者将学会如何在给定两个相同长度的字符串时，判断第二个字符串是否可以通过一系列的乱序操作得到第一个字符串。

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.

这题我不大会...上网查了一下解法原来是用三维DP。

把s1分为两部分，s11和s12，则s2也可以分为两部分s21和s22。如果s11<-->s21 && s12<-->s22，或者s11<-->s22&&s12<-->s21，那么整个string s1和s2也scramble。

用dp[i][j][len]来表示s1以i为开头，s2以j为开头且长度为len的字串是否scramble，那么把s1(i, i+len-1)从中间分成两部分，和s2(j,j+len-1)分成的两部分分别比较是否scramble。

递推公式为dp[i][j][len] |= (dp[i][j][k] && dp[i+k][j+k][len-k]) || (dp[i][j+len-k][k] && dp[i+k][j][len-k])

base case是len为1时看s[i] == s[j]

填充一个三维数组，每次填充需要让k从1到len都走一遍，时间复杂度是O(n^4)

public class Solution {
    public boolean isScramble(String s1, String s2) {
        if(s1.length() != s2.length())
            return false;
        int n = s1.length();
        boolean[][][] dp = new boolean[n][n][n + 1];
        
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                dp[i][j][1] = s1.charAt(i) == s2.charAt(j);
            }
        }
        
        for(int len = 2; len <= n; len++){
            for(int i = 0; i < n - len + 1; i++){
                for(int j = 0; j < n - len + 1; j++){
                    for(int k = 1; k < len; k++){
                        dp[i][j][len] |= (dp[i][j][k] && dp[i + k][j + k][len - k]) || (dp[i][j + len - k][k] && dp[i + k][j][len - k]);
                    }
                }
            }
        }
        return dp[0][0][n];
    }
}