LeetCode 13.7 Scramble String

Given a string s1,
we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = 

"great"

:

great
/    \
gr    eat
/ \    /  \
g   r  e   at
/ \
a   t

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node 

"gr"

 and swap its two children, it produces a scrambled string 

"rgeat"

.

rgeat
/    \
rg    eat
/ \    /  \
r   g  e   at
/ \
a   t

We say that 

"rgeat"

 is a scrambled string of 

"great"

.

Similarly, if we continue to swap the children of nodes 

"eat"

 and 

"at"

, it produces a scrambled string 

"rgtae"

.

rgtae
/    \
rg    tae
/ \    /  \
r   g  ta  e
/ \
t   a

We say that 

"rgtae"

 is a scrambled string of 

"great"

.

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

递归// LeetCode, Interleaving String// 递归,会超时,仅用来帮助理解// 时间复杂度 O(n^6),空间复杂度 O(1)class Solution {public:    bool isScramble(string s1, string s2) {        return isScramble(s1.begin(), s1.end(), s2.begin());    } private:    typedef string::iterator Iterator;    bool isScramble(Iterator first1, Iterator last1, Iterator first2) {        auto length = distance(first1, last1);        auto last2 = next(first2, length);        if (length == 1) return *first1 == *first2;        for (int i = 1; i < length; ++i)            if ((isScramble(first1, first1 + i, first2)                 && isScramble(first1 + i, last1, first2 + i))                || (isScramble(first1, first1 + i, last2 - i) && isScramble(first1 + i, last1, first2)))                return true;        return false;    }};
动规// LeetCode, Interleaving String// 动规,时间复杂度 O(n^3),空间复杂度 O(n^3)class Solution {public:    bool isScramble(string s1, string s2) {        const int N = s1.size();        if (N != s2.size()) return false;        // f
[i][j],表示长度为 n,起点为 s1[i] 和 // 起点为 s2[j] 两个字符串是否互为 scramble  bool f[N + 1]

;        fill_n(&f[0][0][0], (N + 1) * N * N, false);        for (int i = 0; i < N; i++)            for (int j = 0; j < N; j++)                f[1][i][j] = s1[i] == s2[j];        for (int n = 1; n <= N; ++n) {            for (int i = 0; i + n <= N; ++i) {                for (int j = 0; j + n <= N; ++j) {                    for (int k = 1; k < n; ++k) {                        if ((f[k][i][j] && f[n - k][i + k][j + k]) ||(f[k][i][j + n - k] && f[n - k][i + k][j])) {                            f
[i][j] = true;                            break;                        } }                } }        }        return f
[0][0];    }};
递归 + 剪枝// LeetCode, Interleaving String// 递归 + 剪枝// 时间复杂度 O(n^6),空间复杂度 O(1)class Solution {public:    bool isScramble(string s1, string s2) {        return isScramble(s1.begin(), s1.end(), s2.begin());    } private:    typedef string::iterator Iterator;    bool isScramble(Iterator first1, Iterator last1, Iterator first2) {        auto length = distance(first1, last1);        auto last2 = next(first2, length);        if (length == 1) return *first1 == *first2;        // 剪枝,提前返回        int A[26]; // 每个字符的计数器        fill(A, A + 26, 0);        for(int i = 0; i < length; i++) A[*(first1+i)-'a']++; for(int i = 0; i < length; i++) A[*(first2+i)-'a']--; for(int i = 0; i < 26; i++) if (A[i] != 0) return false;        for (int i = 1; i < length; ++i)            if ((isScramble(first1, first1 + i, first2)                 && isScramble(first1 + i, last1, first2 + i))                || (isScramble(first1, first1 + i, last2 - i) && isScramble(first1 + i, last1, first2)))                return true;        return false;    }};
备忘录法// LeetCode, Interleaving String// 递归 +map 做 cache// 时间复杂度 O(n^3),空间复杂度 O(n^3)
class Solution {public:    bool isScramble(string s1, string s2) {        cache.clear();        return isScramble(s1.begin(), s1.end(), s2.begin());    }private:    typedef string::const_iterator Iterator;    map<tuple<Iterator, Iterator, Iterator>, bool> cache;    bool isScramble(Iterator first1, Iterator last1, Iterator first2) {        auto length = distance(first1, last1);        auto last2 = next(first2, length);        if (length == 1) return *first1 == *first2;        for (int i = 1; i < length; ++i)            if ((getOrUpdate(first1, first1 + i, first2)                 && getOrUpdate(first1 + i, last1, first2 + i))                || (getOrUpdate(first1, first1 + i, last2 - i) && getOrUpdate(first1 + i, last1, first2)))                return true;        return false;    }        bool getOrUpdate(Iterator first1, Iterator last1, Iterator first2) {        auto key = make_tuple(first1, last1, first2);        auto pos = cache.find(key);        return (pos != cache.end()) ?        pos->second : (cache[key] = isScramble(first1, last1, first2));    }};
备忘录法typedef string::const_iterator Iterator; typedef tuple<Iterator, Iterator, Iterator> Key; // 定制一个哈希函数namespace std {    template<> struct hash<Key> {        size_t operator()(const Key & x) const {            Iterator first1, last1, first2;            tie(first1, last1, first2) = x;            int result = *first1;            result = result * 31 + *last1;            result = result * 31 + *first2;            result = result * 31 + *(next(first2, distance(first1, last1)-1));            return result;        }    };    }// LeetCode, Interleaving String// 递归 +unordered_map 做 cache,比 map 快 // 时间复杂度 O(n^3),空间复杂度 O(n^3)class Solution {public:    unordered_map<Key, bool> cache;    bool isScramble(string s1, string s2) {        cache.clear();        return isScramble(s1.begin(), s1.end(), s2.begin());    }    bool isScramble(Iterator first1, Iterator last1, Iterator first2) {        auto length = distance(first1, last1);        auto last2 = next(first2, length);        if (length == 1)            return *first1 == *first2;        for (int i = 1; i < length; ++i)            if ((getOrUpdate(first1, first1 + i, first2)                 && getOrUpdate(first1 + i, last1, first2 + i))                || (getOrUpdate(first1, first1 + i, last2 - i)&& getOrUpdate(first1 + i, last1, first2)))                return true;        return false;    }        bool getOrUpdate(Iterator first1, Iterator last1, Iterator first2) {        auto key = make_tuple(first1, last1, first2);        auto pos = cache.find(key);        return (pos != cache.end()) ?        pos->second : (cache[key] = isScramble(first1, last1, first2));    }};