From 72bd8f1c1d91943c425fd5a5f18c5f61708310d9 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Fri, 13 Aug 2021 14:37:42 +0800
Subject: [PATCH] =?UTF-8?q?233:=E6=95=B0=E5=AD=97=201=20=E7=9A=84=E4=B8=AA?=
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 .../leetcode/editor/cn/NumberOfDigitOne.java  | 58 +++++++++++++++++++
 .../leetcode/editor/cn/NumberOfDigitOne.md    | 26 +++++++++
 2 files changed, 84 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/NumberOfDigitOne.java
 create mode 100644 src/main/java/leetcode/editor/cn/NumberOfDigitOne.md

diff --git a/src/main/java/leetcode/editor/cn/NumberOfDigitOne.java b/src/main/java/leetcode/editor/cn/NumberOfDigitOne.java
new file mode 100644
index 0000000..768ca87
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/NumberOfDigitOne.java
@@ -0,0 +1,58 @@
+//给定一个整数 n，计算所有小于等于 n 的非负整数中数字 1 出现的个数。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：n = 13
+//输出：6
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：n = 0
+//输出：0
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 0 <= n <= 2 * 109 
+// 
+// Related Topics 递归 数学 动态规划 
+// 👍 278 👎 0
+
+package leetcode.editor.cn;
+
+//233:数字 1 的个数
+class NumberOfDigitOne {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new NumberOfDigitOne().new Solution();
+        //4
+        System.out.println(solution.countDigitOne(11));
+        //6
+        System.out.println(solution.countDigitOne(13));
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+
+        public int countDigitOne(int n) {
+            int index = 1;
+            int count = 0;
+            while (index<=n){
+                count += (n / (index * 10)) * index + Math.min(Math.max(n % (index * 10) - index + 1, 0), index);
+                index *= 10;
+            }
+            return count;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/NumberOfDigitOne.md b/src/main/java/leetcode/editor/cn/NumberOfDigitOne.md
new file mode 100644
index 0000000..2648bbd
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+++ b/src/main/java/leetcode/editor/cn/NumberOfDigitOne.md
@@ -0,0 +1,26 @@
+<p>给定一个整数 <code>n</code>，计算所有小于等于 <code>n</code> 的非负整数中数字 <code>1</code> 出现的个数。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 13
+<strong>输出：</strong>6
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>n = 0
+<strong>输出：</strong>0
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>0 <= n <= 2 * 10<sup>9</sup></code></li>
+</ul>
+<div><div>Related Topics</div><div><li>递归</li><li>数学</li><li>动态规划</li></div></div>\n<div><li>👍 278</li><li>👎 0</li></div>
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