[leetcode]132. Palindrome Partitioning II

链接：https://leetcode.com/problems/palindrome-partitioning-ii/description/

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.

Example:

Input: "aab"
Output: 1
Explanation: The palindrome partitioning ["aa","b"] could be produced using 1 cut.

Calculate and maintain 2 DP states:

1. pal[i][j] , which is whether s[i..j] forms a pal

2. d[i], which
   is the minCut for s[i..n-1]

Once we comes to a pal[i][j]==true:

• if j==n-1, the string s[i..n-1] is a Pal, minCut is 0, d[i]=0;
• else: the current cut num (first cut s[i..j] and then cut the rest
  s[j+1...n-1]) is 1+d[j+1], compare it to the exisiting minCut num
  d[i], repalce if smaller.

d[0] is the answer.

 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));
            vector<int> d(n);
            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];
        }
    };