leetcode 233: Number of Digit One

Number of Digit One

Total Accepted: 307 Total Submissions: 1853

Given an integer(整数) n, count the total number of digit(数字) 1 appearing in all non-negative(非负的) integers less than or equal to n.

For example:
Given n = 13,
Return 6, because digit 1 occurred in the following numbers: 1, 10, 11, 12, 13.

reference(参考): https://leetcode.com/discuss/44281/4-lines-o-log-n-c-Java-Python

intuitive(直觉的): 每10个数, 有一个个位是1, 每100个数, 有10个十位是1, 每1000个数, 有100个百位是1.  做一个循环, 每次计算单个位上1得总个数(个位,十位, 百位). 

以算百位上1为例子:   假设百位上是0, 1, 和 >=2 三种情况:

    case 1: n=3141092, a= 31410, b=92. 计算百位上1的个数应该为 3141 *100 次.

    case 2: n=3141192, a= 31411, b=92. 计算百位上1的个数应该为 3141 *100 + (92+1) 次.

    case 3: n=3141592, a= 31415, b=92. 计算百位上1的个数应该为 (3141+1) *100 次.

以上三种情况可以用 一个公式概括:

( + ) /  *  + ( %  = ) * ( + );

[CODE]

1. publicclass Solution {  
2.     public countDigitOne(
3.          ones = 
4.         ; m <= n; m *= 
5.              a = n/m, b = n%m;  
6.             ones += (a + 
7.             ) ones += b + 
8.         return ones;  

public class Solution {    public int countDigitOne(int n) {        int ones = 0;        for (long m = 1; m <= n; m *= 10) {            long a = n/m, b = n%m;            ones += (a + 8) / 10 * m;            if(a % 10 == 1) ones += b + 1;        }        return ones;    }}