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Sunday, December 23, 2012

[LeetCode] Interleaving String 解题报告


Given s1s2s3, find whether s3 is formed by the interleaving of s1 and s2.
For example,
Given:
s1 = "aabcc",
s2 = "dbbca",
When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.
» Solve this problem

[解题思路]
第一个想法是merge sort或许可以做。
1:   bool isInterleave(string s1, string s2, string s3) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function    
4:      if(s3.size() != (s1.size() + s2.size()))  
5:      {  
6:        return false;  
7:      }  
8:      int i =0, j= 0, k=0;  
9:      while(i< s1.size() && j< s2.size())  
10:      {  
11:        if(s1[i] == s3[k])  
12:        {  
13:          i ++;  
14:        }  
15:        else if(s2[j] == s3[k])  
16:        {  
17:          j++;  
18:        }  
19:        else  
20:        {  
21:          return false;  
22:        }  
23:        k++;  
24:      }  
25:      while(i< s1.size())  
26:      {  
27:        if(s1[i] == s3[k])  
28:        {  
29:          i++;k++;  
30:        }  
31:        else  
32:        {  
33:          return false;  
34:        }  
35:      }  
36:      while(j<s2.size())  
37:      {  
38:        if(s2[j] == s3[k])  
39:        {  
40:          j++;k++;  
41:        }  
42:        else  
43:        {  
44:          return false;  
45:        }  
46:      }  
47:      return true;  
48:    }  

但是merge sort没法考虑两个字符串的组合顺序问题。当处理{"C","CA", "CAC"}的时候,就不行了。

最后还是得用DP。对于
s1 = a1, a2 ........a(i-1), ai
s2 = b1, b2, .......b(j-1), bj
s3 = c1, c3, .......c(i+j-1), c(i+j)

定义 match[i][j] 意味着,S1的(0, i)和S2的(0,j),匹配与S3的(i+j)
如果 ai == c(i+j), 那么 match[i][j] = match[i-1][j], 等价于如下字符串是否匹配。

s1 = a1, a2 ........a(i-1)
s2 = b1, b2, .......b(j-1), bj
s3 = c1, c3, .......c(i+j-1)

同理,如果bj = c(i+j), 那么match[i][j] = match[i][j-1];

所以,转移方程如下:
Match[i][j]
    =   (s3.lastChar == s1.lastChar) && Match[i-1][j]
      ||(s3.lastChar == s2.lastChar) && Match[i][j-1]
初始条件:
    i=0 && j=0时,Match[0][0] = true;
    i=0时, s3[j] = s2[j], Match[0][j] |= Match[0][j-1]
           s3[j] != s2[j], Match[0][j] = false;

    j=0时, s3[i] = s1[i], Match[i][0] |= Match[i-1][0]
           s3[i] != s1[i], Match[i][0] = false;


[Code]
1:    bool isInterleave(string s1, string s2, string s3) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function    
4:      bool *matchUp = new bool[s2.size() +1];  
5:      bool *matchDown = new bool[s2.size()+1];  
6:      if(s3.size() != (s1.size() + s2.size())) return false;  
7:      //initialize  
8:      matchDown[0] = true;  
9:      for(int i =1; i< s2.size() +1; i++)  
10:      {  
11:        if(s2[i-1] == s3[i-1])  
12:          matchDown[i] |= matchDown[i-1];  
13:        else  
14:          matchDown[i]= false;  
15:      }  
16:      matchUp[0] = true;  
17:      for(int i =1; i< s1.size() +1; i++)  
18:      {  
19:        if(s1[i-1] == s3[i-1])  
20:          matchUp[0] |= matchDown[0];  
21:        else  
22:          matchUp[0]= false;  
23:        for(int j =1;j<s2.size() +1; j++)  
24:        {  
25:          matchUp[j]=false;  
26:          if(s1[i-1] == s3[i+j-1])  
27:          {  
28:            matchUp[j] |= matchDown[j];  
29:          }  
30:          if(s2[j-1] == s3[i+j-1])  
31:          {  
32:            matchUp[j] |= matchUp[j-1];  
33:          }  
34:        }  
35:        bool* temp = matchUp;  
36:        matchUp = matchDown;  
37:        matchDown = temp;        
38:      }  
39:      return matchDown[s2.size()];  
40:    }  

[总结]
代码实现中注意初始条件即可。用二维数组实现也可,只是浪费点空间。

代码中有个bug,程序结束时,忘了删除数组。 应该加上Delete matchUp; Delete MatchDown;  写惯了C#,再用c++,老是往最后的清理工作。
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