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Monday, March 4, 2013

[LeetCode] Palindrome Partitioning II, Solution

Given a string s, partition s such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s.
For example, given s = "aab",
Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut.
» Solve this problem

[Thoughts]
凡是求最优解的,一般都是走DP的路线。这一题也不例外。首先求dp函数,

定义函数
D[i,n] = 区间[i,n]之间最小的cut数,n为字符串长度

 a   b   a   b   b   b   a   b   b   a   b   a
                     i                                  n
如果现在求[i,n]之间的最优解?应该是多少?简单看一看,至少有下面一个解


 a   b   a   b   b   b   a   b   b   a   b   a
                     i                   j   j+1     n

此时  D[i,n] = min(D[i, j] + D[j+1,n])  i<=j <n。这是个二维的函数,实际写代码时维护比较麻烦。所以要转换成一维DP。如果每次,从i往右扫描,每找到一个回文就算一次DP的话,就可以转换为
D[i] = 区间[i,n]之间最小的cut数,n为字符串长度, 则,

D[i] = min(1+D[j+1] )    i<=j <n

有个转移函数之后,一个问题出现了,就是如何判断[i,j]是否是回文?每次都从i到j比较一遍?太浪费了,这里也是一个DP问题。
定义函数
P[i][j] = true if [i,j]为回文

那么
P[i][j] = str[i] == str[j] && P[i+1][j-1];

基于以上分析,实现如下:
1:       int minCut(string s) {  
2:            int len = s.size();  
3:            int D[len+1];  
4:            bool P[len][len];  
5:            //the worst case is cutting by each char  
6:            for(int i = 0; i <= len; i++)   
7:            D[i] = len-i;  
8:            for(int i = 0; i < len; i++)  
9:            for(int j = 0; j < len; j++)  
10:            P[i][j] = false;  
11:            for(int i = len-1; i >= 0; i--){  
12:                 for(int j = i; j < len; j++){  
13:                      if(s[i] == s[j] && (j-i<2 || P[i+1][j-1])){  
14:                           P[i][j] = true;  
15:                           D[i] = min(D[i],D[j+1]+1);  
16:                      }  
17:                 }  
18:            }  
19:            return D[0]-1;  
20:       }  


或者可以考虑使用回溯+剪枝,比如:

1:    int minCut(string s) {  
2:      int min = INT_MAX;  
3:      DFS(s, 0, 0, min);  
4:      return min;  
5:    }  
6:    void DFS(string &s, int start, int depth, int& min)  
7:    {     
8:      if(start == s.size())  
9:      {  
10:        if(min> depth-1)  
11:          min = depth-1;  
12:        return;  
13:      }  
14:      for(int i = s.size()-1; i>=start; i--)  //find the biggest palindrome first
15:      {  
16:        if(isPalindrome(s, start, i))  
17:        {        
18:          DFS(s, i+1, depth+1, min);  
19:        }  
20:        if(min!=INT_MAX)  //if get result, then stop search.
21:          break;  
22:      }  
23:    }  
24:    bool isPalindrome(string &s, int start, int end)  
25:    {  
26:      while(start< end)  
27:      {  
28:        if(s[start] != s[end])  
29:          return false;  
30:        start++; end--;  
31:      }  
32:      return true;  
33:    }  

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