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400. Nth Digit

Description

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 <= 231 - 1

Solutions

• Java
• C++
• Python
• Go
• Javascript
• C#

• class Solution {
      public int findNthDigit(int n) {
          int k = 1, cnt = 9;
          while ((long) k * cnt < n) {
              n -= k * cnt;
              ++k;
              cnt *= 10;
          }
          int num = (int) Math.pow(10, k - 1) + (n - 1) / k;
          int idx = (n - 1) % k;
          return String.valueOf(num).charAt(idx) - '0';
      }
  }
  

• class Solution {
  public:
      int findNthDigit(int n) {
          int k = 1, cnt = 9;
          while (1ll * k * cnt < n) {
              n -= k * cnt;
              ++k;
              cnt *= 10;
          }
          int num = pow(10, k - 1) + (n - 1) / k;
          int idx = (n - 1) % k;
          return to_string(num)[idx] - '0';
      }
  };
  

• class Solution:
      def findNthDigit(self, n: int) -> int:
          k, cnt = 1, 9
          while k * cnt < n:
              n -= k * cnt
              k += 1
              cnt *= 10
          num = 10 ** (k - 1) + (n - 1) // k
          idx = (n - 1) % k
          return int(str(num)[idx])
  
  

• func findNthDigit(n int) int {
  	k, cnt := 1, 9
  	for k*cnt < n {
  		n -= k * cnt
  		k++
  		cnt *= 10
  	}
  	num := int(math.Pow10(k-1)) + (n-1)/k
  	idx := (n - 1) % k
  	return int(strconv.Itoa(num)[idx] - '0')
  }
  

• /**
   * @param {number} n
   * @return {number}
   */
  var findNthDigit = function (n) {
      let k = 1,
          cnt = 9;
      while (k * cnt < n) {
          n -= k * cnt;
          ++k;
          cnt *= 10;
      }
      const num = Math.pow(10, k - 1) + (n - 1) / k;
      const idx = (n - 1) % k;
      return num.toString()[idx];
  };
  
  

• public class Solution {
      public int FindNthDigit(int n) {
          int k = 1, cnt = 9;
          while ((long) k * cnt < n) {
              n -= k * cnt;
              ++k;
              cnt *= 10;
          }
          int num = (int) Math.Pow(10, k - 1) + (n - 1) / k;
          int idx = (n - 1) % k;
          return num.ToString()[idx] - '0';
      }
  }
  

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