leetcode 400. Nth Digit解题报告

leetcode 400. Nth Digit

在线提交网址: https://leetcode.com/problems/nth-digit/

• Total Accepted: 4356
• Total Submissions: 14245
• Difficulty: Easy

Find the n^th 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:3Output:3

Example 2:

Input:11Output: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.

Tags: Math

分析:

1.1位数共有9=9·1个, 1~9;

2.2位数共有90=9·10个, 10~99;

3.3位数共有900=9·10·10个, 100~999;

…

已AC代码:

#include<cstdio>#include<iostream>using namespace std;class Solution {    public:        int findNthDigit(int n) {            int len = 1, base = 1;  // len表示当前数的位数, base表示当前位是个位、百位、千位等...            while (n > 9L * base * len) {                n -= 9 * base * len;                len++;                base *= 10;            }            int curNum = (n - 1)/len + base, digit = 0;   // curNum是含有所找digit的那个数            for (int i = (n - 1) % len; i < len; ++i) {          // 根据偏移量找到所找的数字                digit = curNum % 10;                curNum /= 10;            }            return digit;        }};// 以下为测试 int main() {    Solution sol;    int n;    cin>>n;  // 150    int res = sol.findNthDigit(n);    cout<<res<<" "<<endl;    return 0;}