LeetCode 87. Scramble String

参考链接：[1]https://blog.csdn.net/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];}
};

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