Leetcode 刷题笔记1 动态规划part05

开始完全背包

不同于01背包，完全背包的特色在于元素可以重复拿取， 因此在递归公式和遍历顺序上都有些许不同。

leetcode 518 零钱兑换 ||

在组合方式中所用到的递推公式是dp[ j ] = dp[ j - coins[i]] + dp[ j ]

对于coins[ i ] > j 的情况，for j in range(coin[ i ], amount + 1)不会执行，即实现dp[ i ][j] = dp[ i - 1][j]

class Solution:
    def change(self, amount: int, coins: List[int]) -> int:
        dp = [0] * (amount + 1)
        dp[0] = 1
        for i in range(len(coins)):
            for j in range(coins[i], amount + 1):
                dp[j] += dp[j - coins[i]]
        return dp[amount]

leetcode 377 组合总和 IV

如果求组合数就是外层for循环遍历物品，内层for遍历背包。

如果求排列数就是外层for遍历背包，内层for循环遍历物品。
测试代码：

from typing import List

class Solution:
    # 方式1: 先遍历物品，再遍历背包（求组合数）
    def change_order1(self, amount: int, coins: List[int]) -> int:
        dp = [0] * (amount + 1)
        dp[0] = 1
        paths = [[] for _ in range(amount + 1)]  # 记录路径
        paths[0].append([])  # 初始化空组合

        for coin in coins:  # 遍历物品
            for j in range(coin, amount + 1):  # 遍历背包
                dp[j] += dp[j - coin]
                for path in paths[j - coin]:  # 复制路径
                    paths[j].append(path + [coin])

        # 打印所有组合
        print(f"先遍历物品，再遍历背包（组合数: {dp[amount]}）：")
        for path in paths[amount]:
            print(path)

        return dp[amount]

    # 方式2: 先遍历背包，再遍历物品（求排列数）
    def change_order2(self, amount: int, coins: List[int]) -> int:
        dp = [0] * (amount + 1)
        dp[0] = 1
        paths = [[] for _ in range(amount + 1)]  # 记录路径
        paths[0].append([])  # 初始化空排列

        for j in range(amount + 1):  # 遍历背包
            for coin in coins:  # 遍历物品
                if j - coin >= 0:
                    dp[j] += dp[j - coin]
                    for path in paths[j - coin]:  # 复制路径
                        paths[j].append(path + [coin])

        # 打印所有排列
        print(f"先遍历背包，再遍历物品（排列数: {dp[amount]}）：")
        for path in paths[amount]:
            print(path)

        return dp[amount]

# PyCharm 运行示例
if __name__ == "__main__":
    solution = Solution()
    amount = 5
    coins = [1, 2, 5]

    # 调用两种方法
    result1 = solution.change_order1(amount, coins)
    result2 = solution.change_order2(amount, coins)

    print(f"先遍历物品，再遍历背包（总组合数）：{result1}")
    print(f"先遍历背包，再遍历物品（总排列数）：{result2}")

解题思路：

class Solution:
    def combinationSum4(self, nums: List[int], target: int) -> int:
        dp = [0] * (target + 1)
        dp[0] = 1
        for i in range(1, target + 1):
            for j in range(len(nums)):
                if i - nums[j] >= 0:
                    dp[i] += dp[i - nums[j]]
        return dp[target]

卡码网 57 爬楼梯

分析好重量m和包容量n可解

import sys

def climbing_stairs(n, m):
    dp = [0] * (n + 1)
    dp[0] = 1
    for j in range(1, n + 1):
        for i in range(1, m + 1):
            if j >= i:
                dp[j] += dp[j - i]
    return dp[n]
 
if __name__ == '__main__':
    n, m = list(map(int, input().split()))
    print(climbing_stairs(n, m))