【leetcode】动态规划刷题总结（三）-背包问题

背包问题

有n件物品和一个最多能背重量为w的背包。第i件物品的重量是weight[i]，得到的价值是value[i] 。01背包是每件物品只能用一次；完全背包是每件物品数量无限。求解目标是将哪些物品装入背包里物品价值总和最大。

01背包问题-二维DP数组解法

遍历顺序的选择：只要正序遍历背包容量即可，遍历物品、背包容量的顺序可以颠倒

def test_2_wei_bag_problem1(weight, value, bagweight):
    # dp[i][j]代表背包空间为j的情况下，从下标为0～i的物品里面任意取，能达到的最大价值
    dp = [[0] * (bagweight + 1) for _ in range(len(weight))]

    # 初始化
    for j in range(weight[0], bagweight + 1):
        dp[0][j] = value[0]

    for i in range(1, len(weight)):  # 遍历物品
        for j in range(bagweight + 1):  # 正序遍历背包容量
            if j < weight[i]:
                dp[i][j] = dp[i - 1][j]
            else:
                dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i])

    return dp[len(weight) - 1][bagweight]

if __name__ == "__main__":
    weight = [1, 3, 4]
    value = [15, 20, 30]
    bagweight = 4
    result = test_2_wei_bag_problem1(weight, value, bagweight)
    print(result)

01背包问题-一维DP数组解法

二维DP数组递归公式：dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i])，可以看出第i层可以由第i-1层得出

遍历顺序的选择：需要倒序遍历背包容量（为了保证每个物品只添加1次），且需要先遍历物品，再遍历背包容量

def test_1_wei_bag_problem(weight, value, bagWeight):
    dp = [0] * (bagWeight + 1)
    for i in range(len(weight)):  # 遍历物品
        for j in range(bagWeight, weight[i] - 1, -1):  # 倒序遍历背包容量
            dp[j] = max(dp[j], dp[j - weight[i]] + value[i])
    return dp[bagWeight]


if __name__ == "__main__":
    weight = [1, 3, 4]
    value = [15, 20, 30]
    bagweight = 4
    result = test_1_wei_bag_problem(weight, value, bagweight)
    print(result)

完全背包问题-一维DP数组解法

遍历顺序的选择：完全背包相对01背包，把倒序遍历背包容量改成正序遍历即可（为了保证每个物品可以添加多次）&#x