leetCode_Scramble String

题意：我觉得中文说不清楚了，英文题照抄了~

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.

思路：其实像这种很复杂的 感觉从上往下走走不出来的，要立刻想到用自顶向上的动态规划求解。用s[len][i][j]表示从s1[i]和s2[j]开始,len个字符能不能互换的条件也很简单：从1到len-1把s1和s2分成两段，要么是s1前面和s2前面、s1后面和s2后面能匹配的上，要么就相反。

代码如下：

class Solution {public:    bool isScramble(string s1, string s2) {        if(s1.length()!=s2.length()) return false;        if(s1.length()==0) return true;        if(s1.length()==1) return s1==s2;        int i,j,k,l,t,len=s1.length();        vector<vector<vector<bool>>> s(len+2,vector<vector<bool>>(len+2,vector<bool>(len+2,false)));        for(i=0;i<len;i++)        {            for(j=0;j<len;j++)            {                s[1][i][j]=(s1[i]==s2[j]);            }        }        for(l=2;l<=len;l++)        {            for(i=0;i<len-l+1;i++)            {                for(j=0;j<len-l+1;j++)                {                    for(k=1;k<l;k++)                    {                        if(s[k][i][j]&&s[l-k][i+k][j+k]||s[k][i][j+l-k]&&s[l-k][i+k][j])                        {                            s[l][i][j]=true;                            break;                        }                    }                }            }        }        return s[len][0][0];    }};