动态规划-01背包问题

01 背包问题

#coding=utf-8

# 递归
class Solution(object):
    def knapsack01(self, w, v,c):
        """
        :type w: list
        :type v: list

        """
        length = len(w)
        return self.bestValue(w,v,length-1,c)


    # 用[0...index]的物品，填充容积为c的背包的最大价值
    def bestValue(self,w,v,index,c):

        # 函数的定义说明了初值
        if index < 0 or c <= 0:
            return 0

        res = self.bestValue(w,v,index-1,c)

        if w[index] <= c:
            res = max(res,v[index] + self.bestValue(w,v,index-1,c-w[index]))

        return res

# 记忆化递归
class Solution2(object):
    def knapsack01(self, w, v, c):
        """
        :type w: list
        :type v: list

        """
        length = len(w)

        self.memo= [[-1 for i in range(c+1)] for i in range(length)]
        return self.bestValue(w, v, length - 1, c)

    # 用[0...index]的物品，填充容积为c的背包的最大价值
    def bestValue(self, w, v, index, c):

        if index < 0 or c <= 0:
            return 0


        if self.memo[index][c] != -1:
            return self.memo[index][c]


        res = self.bestValue(w, v, index - 1, c)

        if w[index] <= c:
            res = max(res, v[index] + self.bestValue(w, v, index - 1, c - w[index]))

        self.memo[index][c] = res
        return res





# 动态规划
class Solution3(object):
    def knapsack01(self, w, v, c):


        if not w :
            return 0

        length = len(w)

        memo = [[-1 for i in range(c+1) ] for i in range(length)]

        for i in range(c+1):

            if w[0] <= i:
                memo[0][i] = v[0]
            else:
                memo[0][i] = 0


        for i in range(1,length):
            for j in range(c+1):
                memo[i][j] = memo[i-1][j]

                if w[i] <= j:
                    memo[i][j] = max(memo[i][j] ,v[i] + memo[i-1][j-v[i]])

        return memo[length-1][c]


# 01背包问题的空间优化
# 两行数组
class Solution4(object):
    def knapsack01(self, w, v, c):

        if not w:
            return 0

        length = len(w)

        memo = [[-1 for i in range(c + 1)] for i in range(2)]

        for i in range(c + 1):

            if w[0] <= i:
                memo[0][i] = v[0]
            else:
                memo[0][i] = 0

        for i in range(1, length):
            for j in range(c + 1):
                memo[i%2][j] = memo[(i-1)%2][j]

                if v[i] <= j:
                    memo[i%2][j] = max(memo[i%2][j], v[i] + memo[(i-1)%2][j - v[i]])

        return memo[(length-1)%2][c]



# 01背包问题的空间优化
# 一行数组
class Solution5(object):
    def knapsack01(self, w, v, c):

        if not w:
            return 0

        length = len(w)

        memo = [-1 for i in range(c + 1)]

        for i in range(c + 1):

            if w[0] <= i:
                memo[i] = v[0]
            else:
                memo[i] = 0

        for i in range(1, length):
            for j in range(c,w[i]-1):
                    memo[j] = max(memo[j], v[i] + memo[j - v[i]])

        return memo[c]