【ATT】Palindrome Partitioning II-优快云博客

本文介绍了一种使用动态规划解决回文字符串分区问题的方法，包括如何检查子串是否为回文以及如何计算最少切割次数。

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.

A: 两个DP的结合： dp[i] = min(1+dp[j-1]) 如果s(i,j)是回文的。

同时检查回文也是一个DP

   int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s.empty())
            return 0;
            
        vector<vector<bool>> bPalin;
        
        int i,j;
        
        for(i=0;i<s.size();i++)
            bPalin.push_back(vector<bool>(s.size()));
            
        for(i=0;i<s.size();i++)
            bPalin[i][i] = true;
            
        for(i=s.size()-2;i>=0;i--)
        {
            bPalin[i][i+1] = (s[i]==s[i+1]?true:false);
            for(j=i+2;j<s.size();j++)
            {
                if(s[i]==s[j]&&bPalin[i+1][j-1])
                    bPalin[i][j] = true;
                else bPalin[i][j] = false;
            }
        }
        
        if(bPalin[0][s.size()-1])
            return 0;
        
        vector<int> dp(s.size());
        for(i=0;i<s.size();i++)
        {
            if(bPalin[0][i])
            {
                dp[i] = 0;
                continue;
            }
            int curmin = INT_MAX;
            for(j=1;j<=i;j++)
            {
                if(bPalin[j][i]&&dp[j-1]+1<curmin)
                    curmin = dp[j-1]+1;
            }
            dp[i] = curmin;
        }
        
        return dp[s.size()-1];
    }

转载于:https://www.cnblogs.com/summer-zhou/p/3335796.html