leetcode 87 Scramble String

https://www.cnblogs.com/grandyang/p/4318500.html

1. 递归，每次比较字符串的i个字串是否匹配。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if(s1.size()!=s2.size()) return false;
        if(s1==s2) return true;
        string str1=s1,str2=s2;
        sort(str1.begin(),str1.end());
        sort(str2.begin(),str2.end());
        if(str1!=str2) return false;
        for(int i=1;i<s1.size();++i) {
            string s11=s1.substr(0,i),s12=s1.substr(i);
            string s21=s2.substr(0,i),s22=s2.substr(i);
            if(isScramble(s11,s21)&&isScramble(s12,s22)) return true;
            s21=s2.substr(s2.size()-i);s22=s2.substr(0,s2.size()-i);
            if(isScramble(s11,s21)&&isScramble(s12,s22)) return true;
        }
        return false;
    }
};

2. dp

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