Palindrome Partitioning II

题目地址：点击打开链接

class Solution {public:    int minCut(string s) {        const int n=s.size();        int f[n+1];        bool p[n][n];        fill_n(&p[0][0],n*n,false);        for(int i=0;i<=n;++i)f[i]=n-1-i;        for(int i=n-1;i>=0;--i){            for(int j=i;j<n;++j){                if(s[i]==s[j]&&(j-i<2||p[i+1][j-1])){                    p[i][j]=true;                    f[i]=min(f[i],f[j+1]+1);                }            }        }        return f[0];    }};