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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.

无耻一点的话，可以再问题一的基础上，找结果集中元素最少的，当然就是分割最少的，厚道一些就dp了。

开始觉得很简答，利用问题一中的 判断子串是否是回文串的函数，进行dp，但是果断超时了，看来是否为回文的信息还要记录，就添加个数组，来记录从i 到 j 的字串是否为回文。

class Solution {public:   /*     dp[k] = min{ dp[k], dp[j - 1] + 1} if s[j...k] is palindrom, 0 <=j <= k - 1;     dp[k] = dp[k - 1] + 1，otherwise.   */   int minCut(string s){		vector
   
     dp(s.size() + 1, 0x7f7f7f7f);		/*		bplin[i][i] = true, but it's useless for this problem.		*/		vector
    
     
       > bpalin(s.size(), vector
      
       (s.size(), false));		dp[0] = -1;		for(int i = 0; i < s.size(); ++i)		{		    //we assume s[i] can't make palindrome with items before it.			dp[i + 1] = dp[i] + 1;			//check if s[i] can make palindrome with items before it.			//if so, we change  dp[i + 1].			for(int cur = i - 1; cur >= 0; --cur)				if(s[i] == s[cur] && (i - cur <= 2 || bpalin[cur + 1][i - 1])){					dp[i + 1] = min(dp[i + 1], dp[cur] + 1);					bpalin[cur][i] = true;				}		   		}		return dp[s.size()];	}};
      
     
    
   

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