87. Scramble String(三维动态规划+形似map的数组使用技巧)

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.

1.动态规划：

建立一个dp[i][j][k]的三维数组，表示从s1的i开始，s2的j开始，长度为k的两个字符串是否满足scramble

详情请看 Scramble String -- LeetCode

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

2.切割递归：

自己写的时候想到了map，在discuss里面发现了不用map的非常简洁明了的代码

贴上来记录一下，以便回顾

class Solution {public:    bool isScramble(string s1, string s2) {        if(s1==s2)            return true;                    int len = s1.length();        int count[26] = {0};        for(int i=0; i<len; i++)        {            count[s1[i]-'a']++;            count[s2[i]-'a']--;        }                for(int i=0; i<26; i++)        {            if(count[i]!=0)                return false;        }                for(int i=1; i<=len-1; i++)        {            if( isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i)))                return true;            if( isScramble(s1.substr(0,i), s2.substr(len-i)) && isScramble(s1.substr(i), s2.substr(0,len-i)))                return true;        }        return false;    }};