Leetcode 233: Number of Digit One, leetcodedigit

Leetcode 233: Number of Digit One, leetcodedigit
Number of Digit OneTotal Accepted:307Total 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.

[Idea]

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

Intuitive: each 10 digits, each with 1 bits, every 100 bits, 10 10 digits with 1 bits, every 1000 bits, and 100 bits with 1 bits. create a loop and calculate the total number (one digit, ten digits, and hundreds of digits) of a single bit ).

Example:

Taking the last hundred bits as an example: assume that the last hundred bits is 0, 1, and> = 2:

Case 1: n = 3141092, a = 31410, B = 92. The number of 1 in a hundred bits should be 3141*100 times.

Case 2: n = 3141192, a = 31411, B = 92. The number of 1 in a hundred bits should be 3141*100 + (92 + 1) times.

Case 3: n = 3141592, a = 31415, B = 92. Calculate the number of 1 in a hundred bits (3141 + 1) x 100 times.

The above three cases can be summarized using a formula:

(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;    }}

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