LC. 131. 分割回文串

动态规划保存是否是回文串结果法

class Solution(object):
    def partition(self, s):
        """
        :type s: str
        :rtype: List[List[str]]
        """
        res, sol = [],[]
        n = len(s)
        f = [[True]*n for i in range(n)]
        for i in range(n-1,-1,-1):   #倒序的目的是因为最后一行应该全是正确的，倒数第二行(n-2)开始比较两个字符是否相同，
        #只需比较[n-2][n-1]也就是最后两个字符是否是回文。之后根据状态方程进行迭代比较
            for j in range(i+1 , n):
                f[i][j] = (s[i] == s[j]) and f[i+1][j-1]
        def backtrack(s, begin):
            if begin == n:
                res.append(sol[:])
                return
            for k in range(begin,n):
                if not f[begin][k]:
                    continue
                sol.append(s[begin:k+1])
                backtrack(s,k+1)
                sol.pop()
        backtrack(s,0)
        return res

调用递归判断

class Solution(object):
    def partition(self, s):
        """
        :type s: str
        :rtype: List[List[str]]
        """
        res, sol = [],[]
        n = len(s)
        def reverse(tmp):
            i,j = 0,len(tmp)-1
            while i < j:
                if tmp[i] != tmp[j]:
                    return 0
                i += 1
                j -= 1
            return 1
        def backtrack(s, begin):
            if begin == n:
                res.append(sol[:])
                return
            for k in range(begin,n):
                tmp = s[begin:k+1]
                if reverse(tmp) == 0:
                    continue
                sol.append(tmp)
                backtrack(s,k+1)
                sol.pop()
        backtrack(s,0)
        return res