[LeetCode] Palindrome Partitioning II, Solution

Given a string s[/i], partition s[/i] such that every substring of the partition is a palindrome.
Return the minimum cuts needed for a palindrome partitioning of s[/i].
For example, given s[/i] =

"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:    }