LeetCode:Scramble String

题目链接

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.                                                                                本文地址

递归解法：s2[0...j]是否可以由s1[0...i]转换 isScramble(s2[0...j], s1[0...i])，可以分解成 i 个子问题（i 其实等于j，因为两个字符串长度不一样，肯定不能互相转换）：

( isScramble(s2[0...k], s1[0...k]) &&  isScramble(s2[k+1...j], s1[k+1...i]) ) || ( isScramble(s2[0...k], s1[i-k...i]) &&  isScramble(s2[k+1...j], s1[0...i-k-1]) )，（k = 0,1,2 ... i-1，k相当于字符串的分割点)

只要一个子问题返回ture，那么就表示两个字符串可以转换。代码如下：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        return isScrambleRecur(s1,s2);
    }
    bool isScrambleRecur(string &s1, string &s2)
    {
        string s1cop = s1, s2cop = s2;
        sort(s1cop.begin(), s1cop.end());
        sort(s2cop.begin(), s2cop.end());
        if(s1cop != s2cop)return false;//两个字符串所含字母不同
        if(s1 == s2)return true;
        
        int len = s1.size();
        for(int i = 1; i < len; i++)//分割位置
        {
            string s1left = s1.substr(0, i);
            string s1right = s1.substr(i);
            string s2left = s2.substr(0, i);
            string s2right = s2.substr(i);
            if(isScrambleRecur(s1left, s2left) && isScrambleRecur(s1right, s2right))
                return true;
            s2left = s2.substr(0, len-i);
            s2right = s2.substr(len-i);
            if(isScrambleRecur(s1left, s2right) && isScrambleRecur(s1right, s2left))
                return true;
        }
        return false;
    }
};

动态规划解法：

递归解法有很多重复子问题，比如s2 = rgeat, s1 = great 当我们选择分割点为0时，要解决子问题 isScramble(reat, geat)，再对该子问题选择分割点0时，要解决子问题 isScramble(eat,eat)；而当我们第一步选择1作为分割点时，也要解决子问题 isScramble(eat,eat)。相同的子问题isScramble(eat,eat)就要解决2次。

动态规划用数组来保存子问题，设dp[k][i][j]表示s2从j开始长度为k的子串是否可以由s1从i开始长度为k的子串转换而成，那么动态规划方程如下

1.    初始条件：dp[1][i][j] = (s1[i] == s2[j] ? true : false)
2.    dp[k][i][j] = ( dp[divlen][i][j] && dp[k-divlen][i+divlen][j+divlen] )  ||  ( dp[divlen][i][j+k-divlen] && dp[k-divlen][i+divlen][j] ) (divlen = 1,2,3...k-1, 它表示子串分割点到子串起始端的距离) ，只要一个子问题返回真，就可以停止计算

代码如下：

class Solution {
public:
    bool isScramble(string s1, string s2) {
        //动态规划解法
        if(s1.size() != s2.size())return false;
        const int len = s1.size();
        //dp[k][i][j]表示s2从j开始长度为k的子串是否可以由s1从i开始长度为k的子串转换而成
        bool dp[len+1][len][len];
        //初始化长度为1的子串的dp值
        for(int i = 0; i <= len-1; i++)
            for(int j = 0; j <= len-1; j++)
                dp[1][i][j] = s1[i] == s2[j] ? true : false;
        for(int k = 2; k <= len; k++)//子串的长度
            for(int i = 0; i <= len-k; i++)//s1的起始位置
                for(int j = 0; j <= len-k; j++)//s2的起始位置
                {
                    dp[k][i][j] = false;
                    //divlen表示两个子串分割点到子串起始端的距离
                    for(int divlen = 1; divlen < k && !dp[k][i][j]; divlen++)
                        dp[k][i][j] = (dp[divlen][i][j] && dp[k-divlen][i+divlen][j+divlen])
                            || (dp[divlen][i][j+k-divlen] && dp[k-divlen][i+divlen][j]);
                }
        return dp[len][0][0];
    }
};

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