[LeetCode] 400. Nth Digit

轉自LeetCode

Find the nth digit of the infinite integer sequence 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...

Note:
n is positive and will fit within the range of a 32-bit signed integer (n < 231).

Example 1:

Input:
3

Output:
3

Example 2:

Input:
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.

<Solution>

想法如下(參考資料)

• 第一步，先確定答案會在哪個區間。是在雙位數還是三位數

• 第二步，找到答案所在的數字

• 第三步，找到哪一個位數是所要的答案

code 如下

C++

	class Solution {
	public:
	int findNthDigit(int n) {
	//>> calculate how many digits the number has
	long numCnt = 9;
	int digits = 1;
	while(n - numCnt * digits > 0) {
	n -= numCnt * digits;
	numCnt *= 10;
	++digits;
	}
	//>> find out which number is
	long number = 1;
	for(int i = 1; i < digits; i++) {
	number *= 10;
	}
	int index = n % digits;
	if(index == 0) {
	index = digits;
	}
	number += (index == digits) ? n / digits - 1 : n / digits;
	//>> find the digit
	for(int i = index; i < digits; i++) {
	number /= 10;
	}
	return number % 10;
	}
	};

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