Leetcode Scramble String

Given A string S1, we may represent it as a binary tree by partitioning it to the Non-empty substrings Recursivel Y.

Below is one possible representation of S1 = "great" :

    Great   /      gr    eat/\    /  g   r  E   at           /           a   t

To scramble the string, we are choose any non-leaf node and swap it to the children.

for example, if we choose the Node " GR "  and swap its-children, it produces a scrambled string < Code style= "Font-family:menlo,monaco,consolas, ' Courier New ', monospace; font-size:12.6000003814697px; PADDING:2PX 4px; Color:rgb (199,37,78); Background-color:rgb (249,242,244) ">" 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 is "rgtae" a scrambled string of "great" .

Given strings S1 and S2 of the same length, determine if S2 is a scrambled string of S1 .

Test instructions: asks if S2 is a split string of S1.

Idea: One is recursive search, each time is a string into the length of I and len-i two to wear, and then cross-compare; the second is DP, set DP[LEN][I][J] represents the length of Len at the beginning of the S1 the first position of the S2 J position of the start of Len is satisfied with the condition.

public class Solution {public    Boolean isscramble (string s1, string s2) {        int len1 = S1.length ();    int len2 = S2.length ();    if (len1! = Len2) return false;    if (len1 = = 1) return s1.equals (S2);            Char str1[] = S1.tochararray ();    Char str2[] = S2.tochararray ();    Arrays.sort (STR1);    Arrays.sort (STR2);    if (Arrays.equals (str1, str2) = = false) return false;        for (int i = 1; i < len1; i++) {    String S11 = s1.substring (0, i);    String S12 = s1.substring (i, len1);    String S21 = s2.substring (0, i);    String S22 = s2.substring (i, len1);    if (isscramble (S11, S21) && isscramble (S12, S22)) return true;    S21 = s2.substring (0, len1-i);    S22 = s2.substring (len1-i, len1);    if (isscramble (S11, S22) && isscramble (S12, S21)) return true;    }            return false;    }}

public class Solution {public    Boolean isscramble (string s1, string s2) {      int len = S1.length ();    if (len! = S2.length ()) return false;    if (len = = 0) return false;        Char c1[] = S1.tochararray ();    Char c2[] = S2.tochararray ();        Boolean dp[][][] = new boolean[len+1][len+1][len+1];    for (int i = 0, i < len; i++) for    (int j = 0; J < Len; j + +)    dp[1][i][j] = c1[i] = = C2[j];        for (int k = 2, k <= Len; k++) for     (int i = len-k, i >= 0; i--) for    (int j = len-k; J >= 0; j--) {
   
    boolean flag = false;    for (int m = 1; M <= k &&!flag; m++) {    flag = (Dp[m][i][j] && dp[k-m][i+m][j+m]) | |    (Dp[m][i][j+k-m] && dp[k-m][i+m][j]);    }    DP[K][I][J] = flag;    }            return dp[len][0][0];}    }
   

Leetcode Scramble String