LeetCode Scramble String(动态规划)

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.

题意：给出两个字符串s1和s2，问s2是否是s1的二叉树的打散结构表示 

思路：用dp(len,i,j)表示从s1的i索引，s2的j索引开始，长度为len的子串是否是打散结构

则要么前k个字符相等或者交叉相等，所有以dp(len,i,j)=(dp(k,i,j) && dp(len - k, i + k, j + k)) || (dp(k, i, j + len - k) && (dp(len - k, i + k, j))

具体代码如下：
public class Solution
{
public boolean isScramble(String s1, String s2) {
int len1 = s1.length();
int len2 = s2.length();

if (len1 != len2) return false;

boolean[][][] dp = new boolean[len1 + 1][len1 + 1][len1 + 1];

for (int i = 0; i < len1; i++) {
for (int j = 0; j < len2; j++) {
dp[1][i][j] = s1.charAt(i) == s2.charAt(j);
}
}

for (int len = 2; len <= len1; len++) {
for (int i = 0; i <= len1 - len; i++) {
for (int j = 0; j <= len1 - len; j++) {
for (int k = 1; k < len && !dp[len][i][j]; k++) {
dp[len][i][j] = (dp[k][i][j] && dp[len - k][i + k][j + k]) || (dp[k][i + len - k][j] && dp[len - k][i][j + k]);
}
}
}
}

return dp[len1][0][0];

}
}