[LeetCode] Palindrome Partitioning II （DP）

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.

避免用递归产生低效的程序，用DP的方法：

class Solution {
    public:
        int minCut(string s) {
            if(s.empty()) return 0;
            int n = s.size();
            vector<vector<bool>> pal(n,vector<bool>(n,false));//记录Si到Sj之间是否形成回文
            vector<int> d(n);//记录Si到S末尾分割能形成回文的最小分割数
            for(int i=n-1;i>=0;i--)
            {
                d[i]=n-i-1;
                for(int j=i;j<n;j++)
                {
                    if(s[i]==s[j] && (j-i<2 || pal[i+1][j-1]))
                    {
                       pal[i][j]=true;
                       if(j==n-1)
                           d[i]=0;
                       else if(d[j+1]+1<d[i])
                           d[i]=d[j+1]+1;
                    }
                }
            }
            return d[0];
        }
 };   

每个两两之间都考虑到，需要O（n^2）次数，然后只考虑其中能构成回文的子序列，即Si和Sj之间能构成回文的子序列pal[i][j]=true，仔细体会一下。