Update 400 solution

website/content/ChapterFour/0300~0399/0399.Evaluate-Division.md

@@ -120,5 +120,5 @@ func calcEquation(equations [][]string, values []float64, queries [][]string) []
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website/content/ChapterFour/0400~0499/0400.Nth-Digit.md

@@ -0,0 +1,74 @@
+# [400. Nth Digit](https://leetcode.com/problems/nth-digit/)
+
+## 题目
+
+Given an integer n, return the nth digit of the infinite integer sequence [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...].
+
+**Example 1**:
+
+    Input: n = 3
+    Output: 3
+
+**Example 2**:
+
+    Input: n = 11
+    Output: 0
+    Explanation: The 11th digit of the sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ... is a 0, which is part of the number 10.
+
+**Constraints:**
+
+- 1 <= n <= int(math.Pow(2, 31)) - 1
+
+## 题目大意
+
+给你一个整数 n ，请你在无限的整数序列 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...] 中找出并返回第 n 位数字。
+
+## 解题思路
+
+- bits = 1 的时候有 1,2,3,4,5,6,7,8,9 这 9 个数; 9 = math.Pow10(bits - 1) * bits
+- bits = 2 的时候有 10-99 这 90 个数; 90 = math.Pow10(bits - 1) * bits
+- n 不断减去 bits 从 1 开始的数字总数,求出 n 所在的数字是几位数即 bits
+- 计算 n 所在的数字 num，等于初始值加上 (n - 1) / bits
+- 计算 n 所在这个数字的第几位 digitIdx 等于 (n - 1) % bits
+- 计算出 digitIdx 位的数字
+  
+  ### 以11 为例:
+  11 - 9 = 2 
+  
+  (2 - 1) / 2 = 0
+  
+  (2 - 1) % 2 = 1
+  
+  也就是说第 11 位数字是位数是 2 的第一个数字的第二位，即是 0
+
+## 代码
+
+```go
+
+package leetcode
+
+import "math"
+
+func findNthDigit(n int) int {
+	if n <= 9 {
+		return n
+	}
+	bits := 1
+	for n > 9*int(math.Pow10(bits-1))*bits {
+		n -= 9 * int(math.Pow10(bits-1)) * bits
+		bits++
+	}
+	idx := n - 1
+	start := int(math.Pow10(bits - 1))
+	num := start + idx/bits
+	digitIdx := idx % bits
+	return num / int(math.Pow10(bits-digitIdx-1)) % 10
+}
+```
+
+
+----------------------------------------------
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website/content/ChapterFour/0400~0499/0401.Binary-Watch.md

@@ -178,6 +178,6 @@ func findReadBinaryWatchHour(target, index int, c []int, res *[]string) {
 
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