leetcode 400. Nth Digit

前记
不得不说自己还欠缺很多, 这道easy的leetcode原题就在今年阿里笔试编程题(第二题)出现了, 当时没有AC, 做题不够实在伤感...

Description

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.

My solution

下面代码自己AC代码, 查看discuss发现大家思路都是这样做.

class Solution {
public:
    int findNthDigit(int n) {
        long long sum = 0;
        long long tens = 1;
        int i;
        for (i = 1; sum < n; ++i, tens *= 10) {
            sum += i * tens * 9;
        }
        tens /= 10;
        --i;
        sum -= tens * 9 * i;
        int dif = n - sum;
        int whichNum = dif / i + tens;
        int yu = dif % i;
        if (yu == 0) return (whichNum - 1) % 10;
        while (--yu) tens /= 10;
        whichNum /= tens;
        return whichNum % 10;
    }
};

Reference

• leetcode 400