[Leetcode] [JavaScript] Number of Digit One

Number of Digit One

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.

Maths problem, it's really embarrassing for me to be a math clumsy.

Recursive division, take 8192 to raise chestnuts:

Split the 8192 into:

1-999, recursion (999)

1000-1999-1000 1 + recursion (999)

2001-2999, recursion (999)

.

.

8000-8192, recursion (192)

Total is: recursive (999) *8 + 1000 + recursive (192)

Notice that if it's 1192,

Total: Recursive (999) * + (1000-192 + 1) + recursive (192)

(1000-192 + 1) is 1 of the thousands on the 1000-1192.

1 /**2 * @param {number} n3 * @return {number}4  */5 varCountdigitone =function(n) {6     if(n <= 0){7         return0;8}Else if(N < 10){9         return1;Ten     } One     varLen =n.tostring (). length; A     varBase = Math.pow (Ten, len-1); -     varAnswer = parseint (N/base); -     varremainder = n%Base; the     varOneinbase = 0; -     if(Answer = = 1){ -Oneinbase = n-base + 1; -}Else{ +Oneinbase =Base; -     } +     returnCountdigitone (base-1) * answer + oneinbase +countdigitone (remainder); A};

Then less open a few variables, forced to write code on one line, in the actual project do not do so oh, after a period of time you can not understand.

1 /**2 * @param {number} n3 * @return {number}4  */5 varCountdigitone =function(n) {6     if(n <= 0)return0;7     if(N < 10)return1;8     varBase = Math.pow (Ten, N.tostring (). length-1);9     varAnswer = parseint (N/base);Ten     returnCountdigitone (base-1) * answer + (answer = = = 1?) (N-base + 1): Base) + countdigitone (n%base); One};

[Leetcode] [JavaScript] Number of Digit One