[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.

递归分治，拿8192举栗子：

把8192拆成：

1-999 -> 递归(999)

1000-1999 -> 1000个1 + 递归(999)

2001-2999 -> 递归(999)

.

.

8000-8192 -> 递归(192)

总数是：递归(999)*8 + 1000 + 递归(192)

要注意到如果是1192

总数是：递归(999)*1 + (1000 - 192 + 1) + 递归(192)

(1000 - 192 + 1)是1000-1192中千位上的1。

 1 /**
 2  * @param {number} n
 3  * @return {number}
 4  */
 5 var countDigitOne = function(n) {
 6     if(n <= 0){
 7         return 0;
 8     }else if(n < 10){
 9         return 1;
10     }
11     var len = n.toString().length;
12     var base = Math.pow(10, len - 1);
13     var answer = parseInt(n / base);
14     var remainder = n % base;
15     var oneInBase = 0;
16     if(answer === 1){
17         oneInBase = n - base + 1;
18     }else{
19         oneInBase = base;
20     }
21     return countDigitOne(base - 1) * answer + oneInBase + countDigitOne(remainder);
22 };

然后少开几个变量，强行把代码写在一行上，在实际项目中不要这么做哦，过段时间自己也读不懂了。

 1 /**
 2  * @param {number} n
 3  * @return {number}
 4  */
 5 var countDigitOne = function(n) {
 6     if(n <= 0) return 0;
 7     if(n < 10) return 1;
 8     var base = Math.pow(10, n.toString().length - 1);
 9     var answer = parseInt(n / base);
10     return countDigitOne(base - 1) * answer + (answer === 1 ? (n - base + 1) : base) + countDigitOne(n % base);
11 };