[Leetcode] Scramble String

最新推荐文章于 2019-08-26 13:50:26 发布

最新推荐文章于 2019-08-26 13:50:26 发布 · 100 阅读

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本文介绍了一种判断两个字符串是否可以通过乱序操作相互转换的算法。通过递归和动态规划两种方法实现，前者直观但效率较低，后者利用记忆化搜索优化了重复子问题的计算。

Scramble String

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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.

递归的来搞。比较粗暴。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int n = s1.size();
        if (!isSameChar(s1, s2)) return false;
        if (s1 == s2) return true;
        for (int i = 1; i < n; i++) {
            string s11 = s1.substr(0, i);
            string s12 = s1.substr(i);
            string s21 = s2.substr(0, i);
            string s22 = s2.substr(i);
            if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
            s21 = s2.substr(0, n - i);
            s22 = s2.substr(n - i);
            if (isScramble(s11, s22) && isScramble(s12,s21)) return true;
        }
        return false;
    }
    
    bool isSameChar(string a, string b) {
        int c[256] = {0};
        for (int i = 0; i < a.size(); i++) {
            c[a[i]]++;
        }
        for (int i = 0; i < b.size(); i++) {
            if (--c[b[i]] < 0) return false;
        }
        return true;
    }
};

dp[i][j][k] means : s1[i~i+k], s2[j~j+k] is scramble.

// dp[i][j][k] =

// (dp[i][j][t] && dp[i+t][j+t][k-t]) ||

// (dp[i][j+k-t][t] &&dp[i+t][j][k-t]), (1<= t <= k-1)

Note:

loop need to begin from n - 1 to 0, if not, the result is not correct.

because s[0][0][n] need k from 1~n-1, so s[1][1][n-1] must be in the front of s[0][0][n].

class Solution {
public:
    bool isScramble(string s1, string s2) {
        int n = s1.size();
        bool dp[100][100][100] = {0};
        memset(dp, 0, sizeof(dp));
        // dp[i][j][k] 
        //  if true, s1[i~i+k], s2[j~j+k] match.
        //  else dismatch.
        // dp[i][j][k] = 
        //          (dp[i][j][t] && dp[i+t][j+t][k-t]) || 
        //          (dp[i][j+k-t][t] &&dp[i+t][j][k-t]), (1<= t <= k-1)
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j][0] = true;
                dp[i][j][1] = (s1[i] == s2[j]);
            }
        }

        for (int i = n - 1; i >= 0; i--) {
            for (int j = n - 1; j >= 0; j--) {
                for (int k = 2; (i+k <= n) && (j+k <= n); k++) {
                    for (int t = 1; t < k; t++) {
                        if (dp[i][j][k]) break;
                        dp[i][j][k] |= (dp[i][j][t] && dp[i+t][j+t][k-t]) ||
                                        (dp[i][j+k-t][t] && dp[i+t][j][k-t]); 
                    }
                }
            }
        }
        return dp[0][0][n];;
    }
};