[LintCode] Scramble String 爬行字符串

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 ifs2 is a scrambled string of s1.

Have you met this question in a real interview?

Yes

Example

 

Challenge

O(n3) time

LeetCode上的原题，请参见我之前的博客Scramble String。

解法一：

class Solution {
public:
    /**
     * @param s1 A string
     * @param s2 Another string
     * @return whether s2 is a scrambled string of s1
     */
    bool isScramble(string& s1, string& s2) {
        if (s1 == s2) return true;
        if (s1.size() != s2.size()) return false;
        string t1 = s1, t2 = s2;
        sort(t1.begin(), t1.end());
        sort(t2.begin(), t2.end());
        if (t1 != t2) return false;
        int n = s1.size();
        for (int i = ; i < s1.size(); ++i) {
            string a1 = s1.substr(, i), b1 = s1.substr(i), a2 = s2.substr(, i), b2 = s2.substr(i);
            string a3 = s2.substr(n - i), b3 = s2.substr(, n - i);
            if ((isScramble(a1, a2) && isScramble(b1, b2)) || (isScramble(a1, a3) && isScramble(b1, b3))) {
                return true;
            }
        }
        return false;
    }
};

解法二：

class Solution {
public:
    /**
     * @param s1 A string
     * @param s2 Another string
     * @return whether s2 is a scrambled string of s1
     */
    bool isScramble(string& s1, string& s2) {
        if (s1 == s2) return true;
        if (s1.size() != s2.size()) return false;
        int n = s1.size();
        vector<vector<vector<bool>>> dp(n, vector<vector<bool>>(n, vector<bool>(n + , false)));
        for (int i = n - ; i >= ; --i) {
            for (int j = n - ; j >= ; --j) {
                for (int k = ; k <= n - max(i, j); ++k) {
                    if (s1.substr(i, k) == s2.substr(j, k)) {
                        dp[i][j][k] = true;
                    } else {
                        for (int t = ; t < k; ++t) {
                            if ((dp[i][j][t] && dp[i + t][j + t][k - t]) || (dp[i][j + k - t][t] && dp[i + t][j][k - t])) {
                                dp[i][j][k] = true;
                                break;
                            }
                        }
                    }
                }
            }
        }
        return dp[][][n];
    }
};

解法三：

class Solution {
public:
    /**
     * @param s1 A string
     * @param s2 Another string
     * @return whether s2 is a scrambled string of s1
     */
    bool isScramble(string& s1, string& s2) {
        if (s1 == s2) return true;
        if (s1.size() != s2.size()) return false;
        int n = s1.size(), m[] = {};
        for (int i = ; i < n; ++i) {
            ++m[s1[i] - 'a'];
            --m[s2[i] - 'a'];
        }
        for (int i = ; i < ; ++i) {
            if (m[i] != ) return false;
        }
        for (int i = ; i < n; ++i) {
            string a1 = s1.substr(, i), b1 = s1.substr(i);
            string a2 = s2.substr(, i), b2 = s2.substr(i), a3 = s2.substr(n - i), b3 = s2.substr(, n - i);
            if ((isScramble(a1, a2) && isScramble(b1, b2)) || (isScramble(a1, a3) && isScramble(b1, b3))) {
                return true;
            }
        }
        return false;
    }
};