leetcode 233: Number of Digit One

Number of Digit One

  Total Accepted: 307   Total Submissions: 1853

Given an integer n, count the total number of digit 1 appearing in all non-negative 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.

[思路]

reference: https://leetcode.com/discuss/44281/4-lines-o-log-n-c-java-python

intuitive: 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).  

例子:

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:

(a + 8) / 10 * m + (a % 10 == 1) * (b + 1);

[CODE]

 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;
     }
 }