leetcode:Palindrome Partitioning II

最新推荐文章于 2019-01-13 06:57:42 发布

martin_liang

最新推荐文章于 2019-01-13 06:57:42 发布

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本文介绍了一种算法，用于计算将给定字符串分割成多个回文子串所需的最少分割次数。通过动态规划方法，该算法有效地解决了问题，并给出了具体实现。

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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.

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class Solution {
public:
    int minCut(string s) {
        
        vector<vector<bool>> dp(s.size(), vector<bool>(s.size()));
        
        for (int i=0; i<s.size(); i++)
            dp[i][i] = true;
            
        for (int i=1; i<s.size(); i++)
            if (s[i-1] == s[i])
                dp[i-1][i] = true;
                
        for (int len=2; len<=s.size(); len++)
        {
            //这里等号要注意
            for (int i=0; i<=(int)s.size()-1-len; i++)
            {
                if (s[i] == s[i+len] && dp[i+1][i+len-1])
                    dp[i][i+len] = true;
            }
        }
        
        vector<int> auxDP(s.size()+1);
        auxDP[s.size()] = -1; //这里的初始化为－1 也要注意
        
        for (int i=s.size()-1; i>=0; i--)
        {
            auxDP[i] = INT_MAX;//这里要先赋值
            for (int j=i; j<s.size(); j++)
            {
                if (dp[i][j])
                {
                    auxDP[i] = min(auxDP[i], auxDP[j+1]+1);
                }
            }
        }
        return auxDP[0];  
    }
};