diff --git a/codes/python/chapter_dynamic_programming/knapsack.py b/codes/python/chapter_dynamic_programming/knapsack.py
new file mode 100644
index 000000000..9f8d3cff4
--- /dev/null
+++ b/codes/python/chapter_dynamic_programming/knapsack.py
@@ -0,0 +1,99 @@
+"""
+File: knapsack.py
+Created Time: 2023-07-03
+Author: Krahets (krahets@163.com)
+"""
+
+
+def knapsack_dfs(wgt, val, i, c):
+    """0-1 背包：暴力搜索"""
+    # 若已选完所有物品或背包无容量，则返回价值 0
+    if i == 0 or c == 0:
+        return 0
+    # 若超过背包容量，则只能不放入背包
+    if wgt[i - 1] > c:
+        return knapsack_dfs(wgt, val, i - 1, c)
+    # 计算不放入和放入物品 i 的最大价值
+    no = knapsack_dfs(wgt, val, i - 1, c)
+    yes = knapsack_dfs(wgt, val, i - 1, c - wgt[i - 1]) + val[i - 1]
+    # 返回两种方案中价值更大的那一个
+    return max(no, yes)
+
+
+def knapsack_dfs_mem(wgt, val, mem, i, c):
+    """0-1 背包：记忆化搜索"""
+    # 若已选完所有物品或背包无容量，则返回价值 0
+    if i == 0 or c == 0:
+        return 0
+    # 若已有记录，则直接返回
+    if mem[i][c] != -1:
+        return mem[i][c]
+    # 若超过背包容量，则只能不放入背包
+    if wgt[i - 1] > c:
+        return knapsack_dfs_mem(wgt, val, mem, i - 1, c)
+    # 计算不放入和放入物品 i 的最大价值
+    no = knapsack_dfs_mem(wgt, val, mem, i - 1, c)
+    yes = knapsack_dfs_mem(wgt, val, mem, i - 1, c - wgt[i - 1]) + val[i - 1]
+    # 记录并返回两种方案中价值更大的那一个
+    mem[i][c] = max(no, yes)
+    return mem[i][c]
+
+
+def knapsack_dp(wgt, val, cap):
+    """0-1 背包：动态规划"""
+    n = len(wgt)
+    # 初始化 dp 列表
+    dp = [[0] * (cap + 1) for _ in range(n + 1)]
+    # 状态转移
+    for i in range(1, n + 1):
+        for c in range(1, cap + 1):
+            if wgt[i - 1] > c:
+                # 若超过背包容量，则不选物品 i
+                dp[i][c] = dp[i - 1][c]
+            else:
+                # 不选和选物品 i 这两种方案的较大值
+                dp[i][c] = max(dp[i - 1][c - wgt[i - 1]] + val[i - 1], dp[i - 1][c])
+    return dp[n][cap]
+
+
+def knapsack_dp_comp(wgt, val, cap):
+    """0-1 背包：状态压缩后的动态规划"""
+    n = len(wgt)
+    # 初始化 dp 列表
+    dp = [0] * (cap + 1)
+    # 状态转移
+    for i in range(1, n + 1):
+        # 倒序遍历
+        for c in range(cap, 0, -1):
+            if wgt[i - 1] > c:
+                # 若超过背包容量，则不选物品 i
+                dp[c] = dp[c]
+            else:
+                # 不选和选物品 i 这两种方案的较大值
+                dp[c] = max(dp[c - wgt[i - 1]] + val[i - 1], dp[c])
+    return dp[cap]
+
+
+"""Driver Code"""
+if __name__ == "__main__":
+    wgt = [10, 20, 30, 40, 50]
+    val = [60, 100, 120, 160, 200]
+    cap = 50
+    n = len(wgt)
+
+    # 暴力搜索
+    res = knapsack_dfs(wgt, val, n, cap)
+    print(res)
+
+    # 记忆化搜索
+    mem = [[-1] * (cap + 1) for _ in range(n + 1)]
+    res = knapsack_dfs_mem(wgt, val, mem, n, cap)
+    print(res)
+
+    # 动态规划
+    res = knapsack_dp(wgt, val, cap)
+    print(res)
+
+    # 状态压缩后的动态规划
+    res = knapsack_dp_comp(wgt, val, cap)
+    print(res)
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+# 0-1 背包问题
+
+背包问题是学习动态规划的一个非常好的入门题目，其涉及到“选择与不选择”和“限制条件下的最优化”等问题，是动态规划中最常见的问题形式。
+
+背包问题具有很多变种，例如 0-1 背包问题、完全背包问题、多重背包问题等。在本节中，我们先来学习最简单的 0-1 背包问题。
+
+!!! question
+
+    给定 $n$ 个物品，第 $i$ 个物品的重量为 $wgt[i-1]$ 、价值为 $val[i-1]$ ，现在有个容量为 $cap$ 的背包，请求解在不超过背包容量下背包中物品的最大价值。
+    
+    请注意，物品编号 $i$ 从 $1$ 开始计数，但数组索引从 $0$ 开始计数，因此物品 $i$ 对应重量 $wgt[i-1]$ 和价值 $val[i-1]$ 。
+
+下图给出了一个 0-1 背包的示例数据，背包内的最大价值为 $220$ 。
+
+![0-1 背包的示例数据](knapsack_problem.assets/knapsack_example.png)
+
+接下来，我们仍然先从回溯角度入手，先给出暴力搜索解法；再引入记忆化处理，得到记忆化搜索和动态规划解法。
+
+## 方法一：暴力搜索
+
+0-1 背包问题是一道典型的“选或不选”的问题，0 代表不选、1 代表选。我们可以将 0-1 背包看作是一个由 $n$ 轮决策组成的搜索过程，对于每个物体都有不放入和放入两种决策。不放入背包，背包容量不变；放入背包，背包容量减小。由此可得：
+
+- **状态包括物品编号 $i$ 和背包容量 $c$**，记为 $[i, c]$ 。
+- 状态 $[i, c]$ 对应子问题“**前 $i$ 个物品在容量为 $c$ 背包中的最大价值**”，解记为 $dp[i, c]$ 。
+
+当我们做出物品 $i$ 的决策后，剩余的是前 $i-1$ 个物品的子问题，因此状态转移分为两种：
+
+- **不放入物品 $i$** ：背包容量不变，状态转移至 $[i-1, c]$ ；
+- **放入物品 $i$** ：背包容量减小 $wgt[i-1]$ ，价值增加 $val[i-1]$ ，状态转移至 $[i-1, c-wgt[i-1]]$ ；
+
+上述的状态转移向我们展示了本题的「最优子结构」：**最大价值 $dp[i, c]$ 等于不放入物品 $i$ 和放入物品 $i$ 两种方案中的价值更大的那一个**。由此可推出状态转移方程：
+
+$$
+dp[i, c] = \max(dp[i-1, c], dp[i-1, c - wgt[i-1]] + val[i-1])
+$$
+
+以下是暴力搜索的实现代码，其中包含以下要素：
+
+- **递归参数**：状态 $[i, c]$ ；**返回值**：子问题的解 $dp[i, c]$ 。
+- **终止条件**：当已完成 $n$ 轮决策或背包无剩余容量为时，终止递归并返回价值 $0$ 。
+- **剪枝**：若当前物品重量 $wgt[i - 1]$ 超出剩余背包容量 $c$ ，则只能选择不放入背包。
+
+=== "Java"
+
+    ```java title="knapsack.java"
+    [class]{knapsack}-[func]{knapsackDFS}
+    ```
+
+=== "C++"
+
+    ```cpp title="knapsack.cpp"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "Python"
+
+    ```python title="knapsack.py"
+    [class]{}-[func]{knapsack_dfs}
+    ```
+
+=== "Go"
+
+    ```go title="knapsack.go"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="knapsack.js"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="knapsack.ts"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "C"
+
+    ```c title="knapsack.c"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "C#"
+
+    ```csharp title="knapsack.cs"
+    [class]{knapsack}-[func]{knapsackDFS}
+    ```
+
+=== "Swift"
+
+    ```swift title="knapsack.swift"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "Zig"
+
+    ```zig title="knapsack.zig"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+=== "Dart"
+
+    ```dart title="knapsack.dart"
+    [class]{}-[func]{knapsackDFS}
+    ```
+
+如下图所示，由于每个物品都会产生不选和选两条搜索分支，因此最差时间复杂度为 $O(2^n)$ 。
+
+观察递归树，容易发现其中存在一些「重叠子问题」。而当物品较多、背包容量较大，尤其是当相同重量的物品较多时，重叠子问题的数量将会大幅增多。
+
+![0-1 背包的暴力搜索递归树](knapsack_problem.assets/knapsack_dfs.png)
+
+## 方法二：记忆化搜索
+
+为了防止重复求解重叠子问题，我们借助一个记忆列表 `mem` 来记录子问题的解，其中 `mem[i][c]` 表示前 $i$ 个物品在容量为 $c$ 背包中的最大价值。当再次遇到相同子问题时，直接从 `mem` 中获取记录。
+
+=== "Java"
+
+    ```java title="knapsack.java"
+    [class]{knapsack}-[func]{knapsackDFSMem}
+    ```
+
+=== "C++"
+
+    ```cpp title="knapsack.cpp"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "Python"
+
+    ```python title="knapsack.py"
+    [class]{}-[func]{knapsack_dfs_mem}
+    ```
+
+=== "Go"
+
+    ```go title="knapsack.go"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="knapsack.js"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="knapsack.ts"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "C"
+
+    ```c title="knapsack.c"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "C#"
+
+    ```csharp title="knapsack.cs"
+    [class]{knapsack}-[func]{knapsackDFSMem}
+    ```
+
+=== "Swift"
+
+    ```swift title="knapsack.swift"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "Zig"
+
+    ```zig title="knapsack.zig"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+=== "Dart"
+
+    ```dart title="knapsack.dart"
+    [class]{}-[func]{knapsackDFSMem}
+    ```
+
+引入记忆化之后，所有子问题最多只被计算一次，**因此时间复杂度取决于子问题数量**，也就是 $O(n \times cap)$ 。
+
+![0-1 背包的记忆化搜索递归树](knapsack_problem.assets/knapsack_dfs_mem.png)
+
+## 方法三：动态规划
+
+接下来就是体力活了，我们将“从顶至底”的记忆化搜索代码译写为“从底至顶”的动态规划代码。
+
+=== "Java"
+
+    ```java title="knapsack.java"
+    [class]{knapsack}-[func]{knapsackDP}
+    ```
+
+=== "C++"
+
+    ```cpp title="knapsack.cpp"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "Python"
+
+    ```python title="knapsack.py"
+    [class]{}-[func]{knapsack_dp}
+    ```
+
+=== "Go"
+
+    ```go title="knapsack.go"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="knapsack.js"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="knapsack.ts"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "C"
+
+    ```c title="knapsack.c"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "C#"
+
+    ```csharp title="knapsack.cs"
+    [class]{knapsack}-[func]{knapsackDP}
+    ```
+
+=== "Swift"
+
+    ```swift title="knapsack.swift"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "Zig"
+
+    ```zig title="knapsack.zig"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+=== "Dart"
+
+    ```dart title="knapsack.dart"
+    [class]{}-[func]{knapsackDP}
+    ```
+
+观察下图，动态规划的过程本质上就是填充 $dp$ 列表（矩阵）的过程，时间复杂度也为 $O(n \times cap)$ 。
+
+=== "<1>"
+    ![0-1 背包的动态规划过程](knapsack_problem.assets/knapsack_dp_step1.png)
+
+=== "<2>"
+    ![knapsack_dp_step2](knapsack_problem.assets/knapsack_dp_step2.png)
+
+=== "<3>"
+    ![knapsack_dp_step3](knapsack_problem.assets/knapsack_dp_step3.png)
+
+=== "<4>"
+    ![knapsack_dp_step4](knapsack_problem.assets/knapsack_dp_step4.png)
+
+=== "<5>"
+    ![knapsack_dp_step5](knapsack_problem.assets/knapsack_dp_step5.png)
+
+=== "<6>"
+    ![knapsack_dp_step6](knapsack_problem.assets/knapsack_dp_step6.png)
+
+=== "<7>"
+    ![knapsack_dp_step7](knapsack_problem.assets/knapsack_dp_step7.png)
+
+=== "<8>"
+    ![knapsack_dp_step8](knapsack_problem.assets/knapsack_dp_step8.png)
+
+=== "<9>"
+    ![knapsack_dp_step9](knapsack_problem.assets/knapsack_dp_step9.png)
+
+=== "<10>"
+    ![knapsack_dp_step10](knapsack_problem.assets/knapsack_dp_step10.png)
+
+=== "<11>"
+    ![knapsack_dp_step11](knapsack_problem.assets/knapsack_dp_step11.png)
+
+=== "<12>"
+    ![knapsack_dp_step12](knapsack_problem.assets/knapsack_dp_step12.png)
+
+=== "<13>"
+    ![knapsack_dp_step13](knapsack_problem.assets/knapsack_dp_step13.png)
+
+=== "<14>"
+    ![knapsack_dp_step14](knapsack_problem.assets/knapsack_dp_step14.png)
+
+**接下来考虑状态压缩**。以上代码中的 $dp$ 矩阵占用 $O(n \times cap)$ 空间。由于每个状态都只与其上一行的状态有关，因此我们可以使用两个数组滚动前进，将空间复杂度从 $O(n^2)$ 将低至 $O(n)$ 。代码省略，有兴趣的同学可以自行实现。
+
+那么，我们是否可以仅用一个数组实现状态压缩呢？观察可知，每个状态都是由左上方或正上方的格子转移过来的。假设只有一个数组，当遍历到第 $i$ 行时，该数组存储的仍然是第 $i-1$ 行的状态，为了避免左边区域的格子被覆盖，我们应采取倒序遍历，这样方可实现正确的状态转移。
+
+以下动画展示了在单个数组下从第 $i=1$ 行转换至第 $i=2$ 行的过程。建议你思考一下正序遍历和倒序遍历的区别。
+
+=== "<1>"
+    ![0-1 背包的状态压缩后的动态规划过程](knapsack_problem.assets/knapsack_dp_comp_step1.png)
+
+=== "<2>"
+    ![knapsack_dp_comp_step2](knapsack_problem.assets/knapsack_dp_comp_step2.png)
+
+=== "<3>"
+    ![knapsack_dp_comp_step3](knapsack_problem.assets/knapsack_dp_comp_step3.png)
+
+=== "<4>"
+    ![knapsack_dp_comp_step4](knapsack_problem.assets/knapsack_dp_comp_step4.png)
+
+=== "<5>"
+    ![knapsack_dp_comp_step5](knapsack_problem.assets/knapsack_dp_comp_step5.png)
+
+=== "<6>"
+    ![knapsack_dp_comp_step6](knapsack_problem.assets/knapsack_dp_comp_step6.png)
+
+如以下代码所示，我们仅需将 $dp$ 列表的第一维 $i$ 直接删除，并且将内循环修改为倒序遍历即可。
+
+=== "Java"
+
+    ```java title="knapsack.java"
+    [class]{knapsack}-[func]{knapsackDPComp}
+    ```
+
+=== "C++"
+
+    ```cpp title="knapsack.cpp"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "Python"
+
+    ```python title="knapsack.py"
+    [class]{}-[func]{knapsack_dp_comp}
+    ```
+
+=== "Go"
+
+    ```go title="knapsack.go"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "JavaScript"
+
+    ```javascript title="knapsack.js"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "TypeScript"
+
+    ```typescript title="knapsack.ts"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "C"
+
+    ```c title="knapsack.c"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "C#"
+
+    ```csharp title="knapsack.cs"
+    [class]{knapsack}-[func]{knapsackDPComp}
+    ```
+
+=== "Swift"
+
+    ```swift title="knapsack.swift"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "Zig"
+
+    ```zig title="knapsack.zig"
+    [class]{}-[func]{knapsackDPComp}
+    ```
+
+=== "Dart"
+
+    ```dart title="knapsack.dart"
+    [class]{}-[func]{knapsackDPComp}
+    ```
diff --git a/mkdocs.yml b/mkdocs.yml
index 950a79680..cfcfca309 100644
--- a/mkdocs.yml
+++ b/mkdocs.yml
@@ -216,6 +216,7 @@ nav:
     - chapter_dynamic_programming/index.md
     - 13.1. &nbsp; 初探动态规划（New）: chapter_dynamic_programming/intro_to_dynamic_programming.md
     - 13.2. &nbsp; DP 问题特性（New）: chapter_dynamic_programming/dp_problem_features.md
+    - 13.3. &nbsp; 0-1 背包问题（New）: chapter_dynamic_programming/knapsack_problem.md
   - 14. &nbsp; &nbsp; 附录:
     - 14.1. &nbsp; 编程环境安装: chapter_appendix/installation.md
     - 14.2. &nbsp; 一起参与创作: chapter_appendix/contribution.md