LeetCode-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.

首先通过O(n^2)初始化所有起点终点对是否是回文

判断(i,j)是否是回文可以通过(i+1,j-1)是否是回文来加速

之后通过动归完成，动归复杂度为O(n^2)

class Solution {
public:
    int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s=="")return 0;
        vector<vector<int> > ISP;
        ISP.resize(s.length());
        for(int i=0;i<s.length();i++){
            ISP[i].resize(s.length());
        }
        ISP[0][0]=true;
        for(int i=1;i<s.length();i++){
            ISP[i][i]=true;
            if(s[i-1]==s[i]){
                ISP[i-1][i]=true;
            }
            else{
                ISP[i-1][i]=false;
            }
        }
        for(int j=2;j<s.length();j++){
            for(int i=0;i<s.length()-j;i++){
                if(s[i]==s[i+j]&&ISP[i+1][i+j-1]){
                    ISP[i][i+j]=true;
                }
                else{
                    ISP[i][i+j]=false;
                }
            }
        }
        vector<int> cuts;
        cuts.resize(s.length());
        cuts[0]=0;
        for(int i=1;i<s.length();i++){
            cuts[i]=cuts[i-1]+1;
            if(ISP[0][i])cuts[i]=0;
            else{
                for(int j=1;j<=i;j++){
                    if(ISP[i-j+1][i]&&cuts[i-j]+1<cuts[i]){
                        cuts[i]=cuts[i-j]+1;
                    }
                }
            }
        }
        return cuts[s.length()-1];
    }
};