0-1背包。有 N 件物品，背包能承受的最大重量是 M。总重量不超过背包可承受的最大重量，且价值总和最大。针对这个问题，写出几种由python实现的代码

0-1背包问题是一个经典的动态规划问题，在Python中可以使用递归或迭代的方式来解决。以下是两种常见的实现方法：

**1. 递归解法（Brute Force）：**

def knapsack_recursion(weights, values, capacity, n):
    if n == 0 or capacity == 0:
        return 0
    elif weights[n - 1] > capacity:
        return knapsack_recursion(weights, values, capacity, n - 1)
    else:
        include = values[n - 1] + knapsack_recursion(weights, values, capacity - weights[n - 1], n - 1)
        exclude = knapsack_recursion(weights, values, capacity, n - 1)
        return max(include, exclude)

# 使用示例
weights = [2, 3, 4, 5]
values = [6, 9, 7, 10]
capacity = 8
n = len(weights)
max_value = knapsack_recursion(weights, values, capacity, n)

**2. 动态规划（Dynamic Programming）：**

def knapsack_dp(weights, values, capacity, n):
    dp = [[0 for _ in range(capacity + 1)] for _ in range(n + 1)]
    
    for i in range(1, n + 1):
        for w in range(1, capacity + 1):
            if weights[i - 1] <= w:
                dp[i][w] = max(dp[i - 1][w], values[i - 1] + dp[i - 1][w - weights[i - 1]])
            else:
                dp[i][w] = dp[i - 1][w]

    return dp[n][capacity]

# 使用示例同上
max_value = knapsack_dp(weights, values, capacity, n)

在这两个函数中，`weights`存储物品重量，`values`存储物品价值，`capacity`是背包容量，`n`是物品数。`knapsack_recursion`会遍历所有可能的组合，而`knapsack_dp`则利用了状态转移方程来避免重复计算。