Merge pull request #203 from gostool/leetcode0400

Leetcode0400

leetcode/0400.Nth-Digit/400.Nth Digit.go

@@ -0,0 +1,19 @@
+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
+}

leetcode/0400.Nth-Digit/400.Nth Digit_test.go

@@ -0,0 +1,45 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question400 struct {
+	para400
+	ans400
+}
+
+// para 是参数
+type para400 struct {
+	n int
+}
+
+// ans 是答案
+type ans400 struct {
+	ans int
+}
+
+func Test_Problem400(t *testing.T) {
+
+	qs := []question400{
+
+		{
+			para400{3},
+			ans400{3},
+		},
+
+		{
+			para400{11},
+			ans400{0},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 400------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans400, q.para400
+		fmt.Printf("【input】:%v    【output】:%v\n", p.n, findNthDigit(p.n))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0400.Nth-Digit/README.md

@@ -0,0 +1,66 @@
+# [400. Nth Digit](https://leetcode-cn.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
+}
+```