Scramble String

最新推荐文章于 2021-04-30 19:30:28 发布

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本文探讨了如何判断两个等长字符串s1和s2是否为Scramble关系。通过递归划分s1并对比s2，利用动态规划优化算法效率。

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.

从题目示意图可以看到，如果s1是s2的一个scramble，那必然在s1中存在一个偏移位置i把s1切分成两段 (0,i),( i + 1, s1.length() - 1)这两段，这两段的前后两种组合之一一定能和s2相对应的长度的两段形成符合scramble定义的形式。这里i的取值范围（0, s1.length())如果I等于s1.length()则表示s1没有切分（s1直接等于s2).

对于上边的s1的被切分出来的两段，又可以分别从头开始应用相同的分析逻辑。

到此我们可以看出来这个问题可以分治递归解决，但是同时注意到每层的切分都要遍历当前层所有切分位置，这样必然导致接近分治底层的基本组合情况会被多次计算，这正是动态规划要解决的问题。

最后可以得到下边的动态规划的解法，从底部开始构造，保存中间构造结果，直到把整个串都构造完毕（这时一个迭代解法，用同样的记忆法，也可以用递归来实现）。

class Solution {
public:
    bool isScramble(string s1, string s2) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        int len = s1.length();
        if (len != s2.length()) return false;
        bool ret[len][len][len];
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len; j++) {
                ret[i][j][0] = (s1[i] == s2[j]);
            }
        }

        for (int l = 2; l <= len; l++) {
            for (int i = 0; i <= len - l; i++) {
                for (int j = 0; j <= len - l; j++) {
                    ret[i][j][l - 1] = false;
                    for (int k = 1; k < l; k++) {
                        if ((ret[i][j][k - 1] && ret[i + k][j + k][l - k - 1]) || 
                            (ret[i][j + l - k][k - 1] && ret[i + k][j][l - k - 1])) {
                            ret[i][j][l - 1] = true;
                            break;
                        }
                    }
                }
            }
        }

        return ret[0][0][len - 1];
    }

};