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.

题解：
类似矩阵连乘的动归思路。
dp[i][j]=min(dp[i][k]+dp[k+1][j]+1), i<=k<j.
但是用这个方程的时间是O(n^3)，简化dp[i][j]为dp[i]，表示从0到i的minCut.
dp[i]=min(dp[k]+1(s[k+1, i]是回文串, dp[k]+i-k(s[k+1, i]不是回文串)), 0<=k<i.
这个方法在BOJ上过了，但是在leetcode上面 Judge Large 还是超时。。。
还没找到更快的方法。。。路过的人帮解一下吧。。。
class  Solution {
 public :
     bool  IsPalindrome( string  s,  int  i,  int  j){
         if (i == j)
             return   true ;
         while (i < j){
             if (s[i ++ ] != s[j -- ])
                 return   false ;
        }
         return   true ;
    }
     int  minCut( string  s) {
         //  Start typing your C/C++ solution below
         //  DO NOT write int main() function
         int  n  =  s.length();
         if (n == 0   ||  n == 1 )    
             return   0 ;
        vector < int >  min;
         for ( int  i = 0 ; i < n; i ++ )
            min.push_back( 0 );
         int  tmp, ans;
         for ( int  inter = 1 ; inter < n; inter ++ ){
             if (IsPalindrome(s,  0 , inter)){
                min[inter] = 0 ;
            }
             else {
                ans  =  n + 1 ;
                 for ( int  k = 0 ; k < inter; k ++ ){
                     if (IsPalindrome(s, k + 1 , inter))
                        tmp  =  min[k] + 1 ;
                     else
                        tmp  =  min[k] + inter - k;
                     if (tmp < ans)
                        ans  =  tmp;
                }
                min[inter] = ans;
            }
        }
         return  min[n - 1 ];   
    }
};

经高人hadn't指点，超时是因为判断回文串太慢，加上记忆化数组判断回文串就pass了^-^
class  Solution {
 public :
    vector <  vector < int >   >  map;
     int  IsPalindrome( string   & s,  int  i,  int  j){
         if (i > j)  return   false ;
         if (map[i][j] !=- 1 )
             return  map[i][j];
         if (i == j){
             return  map[i][j]  =   1 ;
        }
            
         if (s[i] != s[j])
             return  map[i][j]  =   0 ;
         else {
             if (j - i == 1 )
                 return  map[i][j]  =   1 ;
             else
                 return  map[i][j]  =  IsPalindrome(s, i + 1 , j - 1 );
        }
            
        
    }
     int  minCut( string  s) {
         //  Start typing your C/C++ solution below
         //  DO NOT write int main() function
         int  n  =  s.length();
         if (n == 0   ||  n == 1 )    
             return   0 ;
        vector < int >  min, vtmp;
        
        min.clear(); vtmp.clear(); map.clear();
         for ( int  i = 0 ; i < n; i ++ ){
            min.push_back( 0 );
            vtmp.push_back( - 1 );
        }
         for ( int  i = 0 ; i < n; i ++ )
            map.push_back(vtmp);
            
         int  tmp, ans;
         for ( int  inter = 1 ; inter < n; inter ++ ){
             if (IsPalindrome(s,  0 , inter)){
                min[inter] = 0 ;
            }
             else {
                ans  =  n + 1 ;
                 for ( int  k = 0 ; k < inter; k ++ ){
                     if (IsPalindrome(s, k + 1 , inter))
                        tmp  =  min[k] + 1 ;
                     else
                        tmp  =  min[k] + inter - k;
                     if (tmp < ans)
                        ans  =  tmp;
                }
                min[inter] = ans;
            }
        }
         return  min[n - 1 ];   
    }
};