【To Understand !!! DP or 递归】LeetCode 87. Scramble String

LeetCode 87. Scramble String

参考链接：[1]https://blog.youkuaiyun.com/makuiyu/article/details/44926439
[2]http://www.cnblogs.com/grandyang/p/4318500.html
Solution1：
递归

class Solution {
public:
    bool isScramble(string s1, string s2) {
        return isScramble(s1, 0, s1.size() - 1, s2, 0, s2.size() - 1);
    }

private:
    bool isScramble(string s1, int a1, int b1, string s2, int a2, int b2) {
        if (b1 - a1 != b2 - a2)
            return false;
        if (a1 == b1)
            return s1[a1] == s2[a2];

        // 剪枝，如果子串中的字符不同，则必然不是Scramble String，不用再进行分割处理了
        int cnt[256] = {0};
        for(int i = a1; i <= b1; ++ i)
            ++cnt[s1[i]];
        for(int i = a2; i <= b2; ++ i)
            --cnt[s2[i]];
        for(int i = 0; i < 256 ;++ i)
            if (cnt[i] != 0)
                return false;

        for (int i = a1; i < b1; ++ i) {
            if (isScramble(s1, a1, i, s2, a2, a2 + i - a1) && 
                isScramble(s1, i + 1, b1, s2, b2 - (b1 - i - 1), b2))
                return true;
            if (isScramble(s1, a1, i, s2, b2 - (i - a1), b2) && 
                isScramble(s1, i + 1, b1, s2, a2, a2 + b1 - i - 1))
                return true;
        }

        return false;
    }
};

Solution2：
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 i = 0; i < n; ++i) {
            for (int j = 0; j < n; ++j) {
                dp[i][j][1] = s1[i] == s2[j];
            }
        }
        for (int len = 2; len <= n; ++len) {
            for (int i = 0; i <= n - len; ++i) {
                for (int j = 0; j <= n - len; ++j) {
                    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;
                        }
                    }
                }
            }
        }
        return dp[0][0][n];
    }
};