LeetCode: Scramble String

这题看了网上答案

class Solution {
public:
    bool isScramble(string s1, string s2) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (s1.size() != s2.size()) return false;
        map<char, int> S1, S2;
        for (int i = 0; i < s1.size(); i++) S1[s1[i]]++;
        for (int i = 0; i < s2.size(); i++) S2[s2[i]]++;
        if (S1 != S2) return false;
        if (s1.size() == 1) return true;
        for (int i = 1; i < s1.size(); i++) {
            bool flag = isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i, s1.size()-i), s2.substr(i, s2.size()-i));
            flag = flag || (isScramble(s1.substr(0, i), s2.substr(s2.size()-i, i)) && isScramble(s1.substr(i, s1.size()-i), s2.substr(0, s2.size()-i)));
            if (flag) return true;
        }
        return false;
    }
};

 三维DP更牛逼！

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

 C#

public class Solution {
    public bool IsScramble(string s1, string s2) {
        if (s1.Length != s2.Length) return false;
        Dictionary<char, int> S1 = new Dictionary<char, int>();
        Dictionary<char, int> S2 = new Dictionary<char, int>();
        for (int i = 0; i < s1.Length; i++) {
            if (S1.ContainsKey(s1[i])) S1[s1[i]]++;
            else S1.Add(s1[i], 1);
        }
        for (int i = 0; i < s2.Length; i++) {
            if (S2.ContainsKey(s2[i])) S2[s2[i]]++;
            else S2.Add(s2[i], 1);
        }
        foreach (var v in S1) {
            if (!S2.ContainsKey(v.Key) || v.Value != S2[v.Key]) return false;
        }
        if (s1.Length == 1) return true;
        for (int i = 1; i < s1.Length; i++) {
            bool flag = IsScramble(s1.Substring(0, i), s2.Substring(0, i)) && IsScramble(s1.Substring(i, s1.Length-i), s2.Substring(i, s2.Length-i));
            flag = flag || (IsScramble(s1.Substring(0, i), s2.Substring(s2.Length-i, i)) && IsScramble(s1.Substring(i, s1.Length-i), s2.Substring(0, s2.Length-i)));
            if (flag) return true;
        }
        return false;
    }
}