Leetcode 400. Nth Digit[easy]

题目：
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.

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继续混题数。
wa了一次，没注意到位数*此位数数字的个数有可能爆int。

class Solution {public:    int findNthDigit(int n) {        return find(1, n);    }    int find(int w, int n) {        int num = max(w) - min(w) + 1;        if ((long long)n > (long long)w * num) return find(w + 1, n - w * num);        else {            int q = (n - 1) / w;            int r = (n - 1) % w;            r++;            n = min(w) + q;            r = w - r;            while (r--) n /= 10;            return n % 10;        }    }    int max(int w) {        int rt = 0;        while (w--) rt = rt * 10 + 9;        return rt;    }    int min(int w) {        int rt = 1;        w--;        while (w--) rt *= 10;        return rt;    }};