LeetCode刷提笔录Scramble String
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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.
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]; }}
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