js 代码面试题

~~~ 统计 1-n 中 1出现的次数。

 

方法一， 转换成字符串，然后通过正则匹配1， 累加次数【或者转换成字符串，在转换成数组，过滤出 1出现的值     ！】

      js:   

function findOne(n) {

Array.from({length: n})

.map((v, i) => i+1)

.reduce((pre, cur) =＞{

const OneCount = String(cur).match(/1/g).?length || 0;

or

const OneCount1 = String(cur).split('').filter(v => v === '1').length;

return pre + OneCount

}, 0)

}

 

method2:  数学方法：将传入的 n 分为 高低位。

 

 

function  findOne2(n, x) {

let factor = 1;

let count = 0;

let  next = parseInt(n / factor);

while (next ! == 0) {

let lower = n - next * factor;

let cur  = n % 10;

let high =  n / (factor * 10);

if (cur > x) {

count += (high + 1) *  factor;

}else if( cur === x) {

count += (high  * factor + lower + 1);

}else {

count +=  high *  factor;

}

factor *= 10;

next = parseInt(n / factor)

 

}

return count

}