From 7507954a2d0a968b8c1b3be41e7d25847a34415b Mon Sep 17 00:00:00 2001
From: CyC2018 <zhengyc101@163.com>
Date: Wed, 28 Oct 2026 13:45:06 +0800
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题目链接改为力扣中的对应题目，题目描述和代码按该题的接口调整。
---
 notes/43. 从 1 到 n 整数中 1 出现的次数.md | 8 ++++----
 1 file changed, 4 insertions(+), 4 deletions(-)

diff --git a/notes/43. 从 1 到 n 整数中 1 出现的次数.md b/notes/43. 从 1 到 n 整数中 1 出现的次数.md
index 89323cc8..2a947e23 100644
--- a/notes/43. 从 1 到 n 整数中 1 出现的次数.md	
+++ b/notes/43. 从 1 到 n 整数中 1 出现的次数.md	
@@ -1,14 +1,14 @@
 # 43. 从 1 到 n 整数中 1 出现的次数
 
-[NowCoder](https://www.nowcoder.com/practice/bd7f978302044eee894445e244c7eee6?tpId=13&tqId=11184&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
+[Leetcode](https://leetcode.cn/problems/1nzheng-shu-zhong-1chu-xian-de-ci-shu-lcof/)
 
 ## 解题思路
 
 ```java
-public int NumberOf1Between1AndN_Solution(int n) {
+public int digitOneInNumber(int num) {
     int cnt = 0;
-    for (int m = 1; m <= n; m *= 10) {
-        int a = n / m, b = n % m;
+    for (long m = 1; m <= num; m *= 10) {
+        long a = num / m, b = num % m;
         cnt += (a + 8) / 10 * m + (a % 10 == 1 ? b + 1 : 0);
     }
     return cnt;