LeetCode OJ Palindrome Partitioning II

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.

还想按照上一题的dfs解法来解，结果超时了，后来在discuss里面看到了更好地解法，用dp：

https://oj.leetcode.com/discuss/6691/my-dp-solution-explanation-and-code

直接摘录了：

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.

代码也是照抄，罪过：

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;  // initialize the d[i] as its max(to a n-length string, the max cut is n - 1)            for(int j=i;j<n;j++)            {                if(s[i]==s[j] && (j-i<2 || pal[i+1][j-1]))  // if a palindrome's two sides are the same of its length is zero                {                   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];    }};