diff --git a/src/main/java/leetcode/editor/cn/W3tCBm.java b/src/main/java/leetcode/editor/cn/W3tCBm.java
new file mode 100644
index 0000000..c3b620e
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/W3tCBm.java
@@ -0,0 +1,76 @@
+//给定一个非负整数 n ，请计算 0 到 n 之间的每个数字的二进制表示中 1 的个数，并输出一个数组。
+//
+//
+//
+// 示例 1:
+//
+//
+//输入: n = 2
+//输出: [0,1,1]
+//解释:
+//0 --> 0
+//1 --> 1
+//2 --> 10
+//
+//
+// 示例 2:
+//
+//
+//输入: n = 5
+//输出: [0,1,1,2,1,2]
+//解释:
+//0 --> 0
+//1 --> 1
+//2 --> 10
+//3 --> 11
+//4 --> 100
+//5 --> 101
+//
+//
+//
+//
+// 说明 :
+//
+//
+// 0 <= n <= 10⁵
+//
+//
+//
+//
+// 进阶:
+//
+//
+// 给出时间复杂度为 O(n*sizeof(integer)) 的解答非常容易。但你可以在线性时间 O(n) 内用一趟扫描做到吗？
+// 要求算法的空间复杂度为 O(n) 。
+// 你能进一步完善解法吗？要求在C++或任何其他语言中不使用任何内置函数（如 C++ 中的 __builtin_popcount ）来执行此操作。
+//
+//
+//
+//
+// 注意：本题与主站 338 题相同：https://leetcode-cn.com/problems/counting-bits/
+// Related Topics 位运算 动态规划 👍 28 👎 0
+
+package leetcode.editor.cn;
+
+//剑指 Offer II 003:前 n 个数字二进制中 1 的个数
+public class W3tCBm {
+ public static void main(String[] args) {
+ Solution solution = new W3tCBm().new Solution();
+ // TO TEST
+ }
+
+ //leetcode submit region begin(Prohibit modification and deletion)
+ class Solution {
+ public int[] countBits(int n) {
+ int[] nums = new int[n + 1];
+ for (int i = 0; i <= n; i++) {
+ for (int j = 0; j < 32; j++) {
+ nums[i] += (i >> j) & 1;
+ }
+ }
+ return nums;
+ }
+ }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
diff --git a/src/main/java/leetcode/editor/cn/doc/all.json b/src/main/java/leetcode/editor/cn/doc/all.json
index 1645c7f..1f4ca08 100644
--- a/src/main/java/leetcode/editor/cn/doc/all.json
+++ b/src/main/java/leetcode/editor/cn/doc/all.json
@@ -1 +1 @@
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\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/content/W3tCBm.md b/src/main/java/leetcode/editor/cn/doc/content/W3tCBm.md
new file mode 100644
index 0000000..35859ea
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/W3tCBm.md
@@ -0,0 +1,51 @@
+给定一个非负整数 n ，请计算 0 到 n 之间的每个数字的二进制表示中 1 的个数，并输出一个数组。
+
+
+
+示例 1:
+
+
+输入: n = 2
+输出: [0,1,1]
+解释:
+0 --> 0
+1 --> 1
+2 --> 10
+
+
+示例 2:
+
+
+输入: n = 5
+输出: [0,1,1,2,1,2]
+解释:
+0 --> 0
+1 --> 1
+2 --> 10
+3 --> 11
+4 --> 100
+5 --> 101
+
+
+
+
+说明 :
+
+
+
+
+
+进阶:
+
+
+ - 给出时间复杂度为
O(n*sizeof(integer)) 的解答非常容易。但你可以在线性时间 O(n) 内用一趟扫描做到吗？
+ - 要求算法的空间复杂度为
O(n) 。
+ - 你能进一步完善解法吗？要求在C++或任何其他语言中不使用任何内置函数（如 C++ 中的
__builtin_popcount ）来执行此操作。
+
+
+
+
+注意：本题与主站 338 题相同：https://leetcode-cn.com/problems/counting-bits/
+
👍 28👎 0
\ No newline at end of file