Submission Details [leetcode] 算法的改进

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最先看到这一题，直觉的解法就是len从1到s1.size()-1，递归调用比较s1和s2长度为len的子串是否相等，以及剩余部分是否相等。

将s1分成s1[len + rest]，分别比较s2[len + rest]和s2[rest + len]

代码如下：

    bool isScramble(string s1, string s2) {return find(s1, s2);    }bool find(string s1, string s2) {if (s1.compare(s2) == 0) return true;for (int i = 1; i < s1.size(); i++){if (find(s1.substr(0, i), s2.substr(0,i)) && find(s1.substr(i), s2.substr(i)))return true;if (find(s1.substr(0, i), s2.substr(s2.size() - i)) && find(s1.substr(i), s2.substr(0, s2.size() - i)))return true;}return false;    }

是一个TLE的算法

改进点1：在find中查看s1和s2是否字母相同，如果不同，返回false

代码如下：

    bool isScramble(string s1, string s2) {return find(s1, s2);    }bool find(string s1, string s2) {if (!haveSameChar(s1, s2)) return false;if (s1.compare(s2) == 0) return true;for (int i = 1; i < s1.size(); i++){if (find(s1.substr(0, i), s2.substr(0,i)) && find(s1.substr(i), s2.substr(i)))return true;if (find(s1.substr(0, i), s2.substr(s2.size() - i)) && find(s1.substr(i), s2.substr(0, s2.size() - i)))return true;}return false;    }bool haveSameChar(string& s1, string& s2){int chars[26] = {0};for (int i = 0; i < s1.size(); i++){chars[s1[i] - 'a'] ++;}for (int i = 0; i < s2.size(); i++){if (chars[s2[i] - 'a'] == 0)  return false;chars[s2[i] - 'a'] --;}return true;}

改进点2：存储状态，将问题转换为三维DP

dp(i, j, l)：s1[i...i+l]和s2[j...j+l]是否满足要求

dp(i, j, l) = dp(i, j, k) && dp(i + k, j + k, l - k) || dp(i, j + l - k, k) && dp(i + k, j, l - k)

    vector<vector<vector<int>>> dp;    bool isScramble(string s1, string s2) {        int size = s1.size();        dp = vector<vector<vector<int>> > (size, vector<vector<int>>(size, vector<int>(size + 1)));        return find(s1, s2, 0, 0, size);    }        bool find(string & s1, string & s2, int l1, int l2, int length)    {        if (dp[l1][l2][length] != 0)            return dp[l1][l2][length] == 1;        dp[l1][l2][length] = -1;        if (length == 1)            dp[l1][l2][length] = ((s1[l1] == s2[l2]) ? 1 : -1);        for (int len = 1; len < length; len++)        {            if (find(s1, s2, l1, l2, len) && find(s1, s2, l1 + len, l2 + len, length - len))            {                dp[l1][l2][length] = 1;                break;            }            if (find(s1, s2, l1, l2 + length - len, len) && find(s1, s2, l1 + len, l2, length - len))            {                dp[l1][l2][length] = 1;                break;            }        }        return dp[l1][l2][length] == 1;    }

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