LC87. Scramble String
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.
思路是对于对于s1，在每个可能分割的位置分割，递归检查是否有可能是同一字符串，basecase是对于length == 1的字符串，s1的s2子串相等。
public class Solution {
public boolean isScramble(String s1, String s2) {
// partialize s1 and s2, if any union of two part can match, return true
if(s1.equals(s2)) return true;
char[] arr1 = s1.toCharArray();
char[] arr2 = s2.toCharArray();
int[] count = new int[26];
for(int i = 0; i < arr1.length; i++){
count[arr1[i] - 'a']++;
count[arr2[i] - 'a']--;
}
for(int i = 0; i < 26; i++) if(count[i] != 0) return false;
for(int i = 1; i < arr1.length; i++){
if(isScramble(s1.substring(i), s2.substring(i)) && isScramble(s1.substring(0, i), s2.substring(0, i))) return true;
if(isScramble(s1.substring(0, i), s2.substring(arr1.length - i)) && isScramble(s1.substring(i), s2.substring(0, arr1.length - i))) return true;
}
return false;
}
}
Quite a few people here said recursion (without cache) is better than dp, and gave a reason that recursion has something called “pruning”. Pruning means nothing when the two strings mismatch.
BTW, recursive (without cache) is O(a^n), dp is O(n^a)