LeetCode Scramble String

题目：

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.

分析及代码参考：http://cozilla.iteye.com/blog/1950242

class Solution {public://DP,dp[i][j][k]表示从s1[i]开始长度为k的字符串与从s2[j]开始长度为k的字符串是否满足scramble//dp[i][j][k] = (dp[i][j][t]&&dp[i+t][j+t][k-t]) || (dp[i][j+k-t][t]&&dp[i+t][j][k-t])，其中1<=t<k;bool isScramble(string s1, string s2) {int n = s1.size();bool dp[100][100][100] = { false };//初始化for (int i = 0; i < n; i++) {for (int j = 0; j < n; j++) {dp[i][j][0] = true;dp[i][j][1] = (s1[i] == s2[j]);}}for (int i = n-1; i >=0; i--) {for (int j = n-1; j >=0; j--) {for (int k = 2; (i + k <= n) && (j + k <= n); k++) {for (int t = 1; t < k; t++) {if (dp[i][j][k])break;elsedp[i][j][k] = (dp[i][j][t] && dp[i + t][j + t][k - t]) || (dp[i][j + k - t][t] && dp[i + t][j][k - t]);}}}}return dp[0][0][n];}};