动态规划_leetcode279(动归版)

标签：记忆化   col   square   递归   return   turn   规划   print   utf-8   

#coding=utf-8

# 递归版
class Solution1(object):
    def numSquares(self, n):
        """
        :type n: int
        :rtype: int
        """

        return self.squareNum(n)


    def squareNum(self,n):

        if n == 0:
            return 0

        if n == 1:
            return 1

        res = n

        for i in range(1,n):
            if n - i ** 2 >= 0:
                res = min(res, 1 + self.squareNum(n- i ** 2))


        return res

# s = Solution1()
#
# print  s.numSquares(13)

# 记忆化递归
class Solution2(object):
    def numSquares(self, n):
        """
        :type n: int
        :rtype: int
        """

        self.memo = [-1 for i in range(n+1)]

        return self.squareNum(n)


    def squareNum(self,n):

        if n == 0:
            self.memo[0] = 0
            return 0

        if n == 1:
            self.memo[1] = 1
            return 1

        res = n

        for i in range(1,n):
            if n - i ** 2 >= 0:
                if self.memo[n- i ** 2] != -1:
                    res = min(res, 1 + self.memo[n- i ** 2])
                else:
                    self.memo[n - i ** 2] = self.squareNum(n - i ** 2)
                    res = min(res, 1 + self.memo[n- i ** 2])


        return res

# s = Solution2()
#
# print  s.numSquares(40)

# 动归
class Solution3(object):
    def numSquares(self, n):

        return self.squareNum(n)

    def squareNum(self, n):

        self.memo = [n for i in range(n + 1)]

        self.memo[0] = 0
        self.memo[1] = 1

        # #memo[2]
        # for i in range(1, 2):
        #     if n - i ** 2 >= 0:
        #         self.memo[2] = min(self.memo[2],1+self.memo[n - i ** 2])
        #
        #
        # #memo[3]
        # for i in range(1, 3):
        #     if n - i ** 2 >= 0:
        #         self.memo[3] = min(self.memo[3],1+self.memo[n - i ** 2])



        for i in range(1,n+1):
            for j in range(1,i):
                if i - j ** 2 >= 0:
                    self.memo[i] = min(self.memo[i], 1 + self.memo[i - j ** 2])


        return self.memo[n]



s = Solution3()

print  s.numSquares(7691)

动态规划_leetcode279(动归版)

标签：记忆化   col   square   递归   return   turn   规划   print   utf-8   

原文地址：https://www.cnblogs.com/lux-ace/p/10546548.html