LeetCode: Palindrome Partitioning II [132]

class Solution {
public:
    int minCut(string s) {
        int len=s.length();
        if(len==0||len==1)return 0;
        //构造回文判别矩阵
        vector<vector<bool> > isPal(len, vector<bool>(len, false));
        //初始化isPal[i][i]
        for(int i=0; i<len; i++)
            isPal[i][i]=true;
        //初始化isPal[i][i+1]
        for(int i=0; i<len-1; i++)
            if(s[i]==s[i+1])isPal[i][i+1]=true;
        //初始化其他i<j的子串
        for(int i=len-3; i>=0; i--)
            for(int j=i+2; j<len; j++)
                isPal[i][j]=(s[i]==s[j])&&isPal[i+1][j-1];
                
        //求最小切割
        vector<int> minCuts(len, 0);  //记录以i结尾的子串的最小切分数
        for(int i=0; i<len; i++){
            int mincut=INT_MAX;
            for(int j=-1; j<i; j++){
                if(isPal[j+1][i]){
                    if(j==-1)mincut=0;
                    else if(minCuts[j]+1<mincut)mincut=minCuts[j]+1;
                }
            }
            if(mincut!=INT_MAX)minCuts[i]=mincut;
        }
        return minCuts[len-1];
    }
};