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Leetcode--Palindrome Partitioning II

Problem Description:

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.

分析:设cut[i] = 区间[0,i]之间最小的cut数,n为字符串长度, 则,
cut[i] = min(cut[i],1+cut[j] )    0<=j<i
有个转移函数之后,一个问题出现了,就是如何判断[j,i]是否是回文?每次都从i到j比较一遍?太浪费了,这里也是一个DP问题。
定义函数
flag[i][j] = true if [i,j]为回文
那么
flag[i][j] = str[i] == str[j] && P[i+1][j-1];

class Solution {
public:

    int minCut(string s) {
        //if(s.size()==0)
        //    return 0;
        int n=s.size();
        vector<vector<int> > flag(n,vector<int>(n,0));
        vector<int> cut(n+1);
        
        for(int i=0;i<=n;i++)
            cut[i]=i-1;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<=i;j++)
            {
                if(s[i]==s[j]&&(i-j<2||flag[j+1][i-1]==1))
                {
                    flag[j][i]=1;
                    cut[i+1]=min(cut[i+1],cut[j]+1);
                }
            }
        }
        return cut[n];
        
    }
};


Leetcode--Palindrome Partitioning II