[LeetCode]Palindrome Partitioning II

最新推荐文章于 2019-10-13 19:23:13 发布

原创最新推荐文章于 2019-10-13 19:23:13 发布 · 1.5k 阅读

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本文介绍了一种使用C++实现的最小切割字符串算法，通过动态规划和回文子串判断，有效地解决了字符串分割问题。

class Solution {
public:
	int minCut(string s) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int len = s.size();
		vector<int> dp(len+1);//dp[i]: minimum cut between [i,len-1]
		for(int i = 0; i <= len; ++i)
			dp[i] = len-i;
		vector<vector<bool>> p;//p[i][j]: sub string [i,j] palindrome? 
		p.resize(len);
		for(int i = 0; i < len; ++i)
			p[i].assign(len, false);
		for (int i = len-1; i >= 0; --i)
		{
			for (int j = i; j < len; ++j)
			{
				if(s[i] == s[j] && (j-i < 2 || p[i+1][j-1]))
				{
					p[i][j] = true;
					dp[i] = min(dp[i], 1+dp[j+1]);
				}
			}
		}
		return dp[0]-1;
	}
};

second time

class Solution {
public:
    int minCut(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int n = s.size();
        if(n == 0) return 0;
        vector<int> f(n+1, INT_MAX);
        f[0] = -1;
        vector<vector<bool> > p(n, vector<bool>(n, false));
        for(int i = 0; i < n; ++i) p[i][i] = true;
        
        for(int i = 1; i <= n; ++i)
        {
            for(int k = 0; k < i; ++k)
            {
                if(s[k] == s[i-1]) p[k][i-1] = (i-2 <= k+1) || p[k+1][i-2];
                if(p[k][i-1]) f[i] = min(f[i], f[k]+1);
            }
        }
        return f[n];
    }
};