LeetCode 87 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.

对于一个字符串，如果可以通过一系列的，“确定任一点（任意一个字符位置）后，前后子字符串交换得到新字符串”的过程变换得到一个新的字符串，那么就认为这两个字符串Scramble
。

Runtime: 3
ms  beats 97.62% of java submissions.

public static  boolean isScramble(String s1, String s2) {
char[] v1 = s1.toCharArray();
char[] v2 = s2.toCharArray();
return isScramble(v1, 0, v1.length - 1, v2, 0, v2.length - 1);
}

private static boolean isScramble(char[] v1, int start1, int end1, char[] v2, int start2, int
end2) {
int[] letters = new int[26];
boolean isSame = true;
for (int i = start1, j = start2; i <= end1; i++, j++) {
letters[v1[i] -'a']++;
letters[v2[j] -'a']--;
isSame = isSame && v1[i] == v2[j];
}
if (isSame) return true;
for (int i = 0; i < 26; i++) if (letters[i] != 0) return false;
for (int i = start1, j = start2; i < end1; i++, j++) {
if (isScramble(v1, start1, i, v2, start2, j)
&& isScramble(v1, i + 1, end1, v2, j + 1, end2)) return true;
if (isScramble(v1, start1, i, v2, end2 - j + start2, end2)
&& isScramble(v1, i + 1, end1, v2, start2, end2 - j + start2 - 1)) return true;
}
return false;
}

https://discuss.leetcode.com/topic/50618/java-2ms-solution-beats-95

Runtime: 4 ms  beats 89.62% of java submissions.
public boolean isScramble(String s1, String s2) {
if (s1.equals(s2)) return true;

int[] letters = new int[26];
for (int i=0; i<s1.length(); i++) {
letters[s1.charAt(i)-'a']++;
letters[s2.charAt(i)-'a']--;
}
for (int i=0; i<26; i++) if (letters[i]!=0) return false;

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

https://discuss.leetcode.com/topic/19158/accepted-java-solution/2