87. Scramble String

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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 ifs2 is a scrambled string of s1.

分析:动态规划问题.判断从s1的第i个字符和从s2的第j个字符开始,长度为len的字符串是不是scramble字符串,其值:

<span style="font-size:14px;"> for(int k=1;k<len;k++){                        flag[i][j][len]|=(flag[i][j][k]&&flag[i+k][j+k][len-k])||(flag[i][j+len-k][k]&&flag[i+k][j][len-k]);                    }</span>

只要有一种情况是true,则其就是scramble串.当len=1时可以通过判断对应字符是否相等得到.

<span style="font-size:14px;">public class Solution {    public boolean isScramble(String s1, String s2) {        int n=s1.length();        boolean[][][] flag=new boolean[n][n][n+1];        for(int i=0;i<n;i++){            for(int j=0;j<n;j++){                flag[i][j][1]=s1.charAt(i)==s2.charAt(j);            }        }                for(int len=2;len<=n;len++){            for(int i=0;i<n-len+1;i++){                for(int j=0;j<n-len+1;j++){                    for(int k=1;k<len;k++){                        flag[i][j][len]|=(flag[i][j][k]&&flag[i+k][j+k][len-k])||(flag[i][j+len-k][k]&&flag[i+k][j][len-k]);                    }                }            }        }        return flag[0][0][n];    }}</span>

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