leetcode[132]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.

class Solution {
public:
    int minCut(string s) {
        if(s.empty())return 0;
        int n=s.length();
        bool **T=new bool*[n];
        for(int i=0;i<n;i++)
        {
            T[i]=new bool[n];
            for(int j=0;j<n;j++)
            T[i][j]=false;
        }
        int *cut=new int[n+1];
        cut[n]=0;
        for(int i=n-1;i>=0;i--)
        {
            cut[i]=n-i;
            for(int j=i;j<n;j++)
            {
                if(s[i]==s[j]&&(i==j||i+1==j||T[i+1][j-1]))
                {
                    T[i][j]=true;
                    cut[i]=cut[i]<1+cut[j+1]?cut[i]:1+cut[j+1];
                }
            }
        }
        return cut[0]-1;
    }

/**
void judge(string s,bool ** &T)
{
    if(s.empty())return;
    for(int i=0;i<s.length();i++)
    {
        T[i][i]=true;
        int left=i-1,right=i;
        while(left>=0&&right<s.length()&&s[left]==s[right])
        {
            T[left--][right++]=true;
        }
        left=i-1,right=i+1;
        while(left>=0&&right<s.length()&&s[left]==s[right])
        {
            T[left--][right++]=true;
        }
    }
}
    int minCut(string s) {
        if(s.empty())return 0;
        int n=s.length();
        bool **T=new bool*[n];
        for(int i=0;i<n;i++)
        {
            T[i]=new bool[n];
            for(int j=0;j<n;j++)
            T[i][j]=false;
        }
        judge(s,T);
        int *cut=new int[n+1];
        cut[n]=0;
        for(int i=n-1;i>=0;i--)
        {
            cut[i]=n-i;
            for(int j=i;j<n;j++)
            {
                if(T[i][j])
                {
                    cut[i]=cut[i]<1+cut[j+1]?cut[i]:1+cut[j+1];
                }
            }
        }
        return cut[0]-1;
    }
*/
};