Palindrome Partitioning II Leetcode Python

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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.

这题的标准解法是用动态规划，我自己先用了dfs的方法但是解出来超时。解法和palindrome I的一样。

class Solution:    def check(self,s):        for index in range(len(s)):            if s[index]!=s[len(s)-index-1]:                return False        return True        def dfs(self,s,count,valuelist):        if len(s)==0:            #solution.append(valuelist)            Solution.minval=min(Solution.minval,count)        for index in range(1,len(s)+1):            if self.check(s[:index])==True:                self.dfs(s[index:],count+1,valuelist+[s[:index]])    def solve(self,s):        Solution.minval=10000        self.dfs(s,0,[])        print Solution.minvaldef main():    s="aazbcc"    solution=Solution()    solution.solve(s)if __name__=="__main__":    main()

做法就是当确认最后都分解好的时候计算所用的迭代次数。

另外一种解法是用动态规划的方法借鉴了 http://www.cnblogs.com/zuoyuan/p/3758783.html

class Solution:    # @param s, a string    # @return an integer   def minCut(self, s):        dp=[0 for i in range(len(s)+1)]        p=[[False for i in range(len(s))] for j in range(len(s))]        for i in range(len(s)+1):            dp[i]=len(s)-i        for i in reversed(range(len(s)-1)):            for j in range(i,len(s)):                if s[i]==s[j] and (j-i<2 or p[i+1][j-1]):                    p[i][j]=True                    dp[i]=min(dp[i],1+dp[j+1])        return dp[0]-1

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