丑数2之自己挖坑埋自己

今天做了丑数2,遇到个坑,卡主自己好久,所以make先,以后有空再仔细研究

先上题目:

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Write a program to find the n-th ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note:
1 is typically treated as an ugly number.
n does not exceed 1690.

先给出我最终通过的代码:

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var nthUglyNumber = function(n) {
//debugger
var arry = []

var arry_two = [2]
var i = 0
var arry_three = [3]
var j = 0
var arry_five = [5]
var k = 0

var a = arry_two[i]
var b = arry_three[i]
var c = arry_five[i]

if (n < 5) {
return n
}
while (1) {
if ( a < b && a < c) {
arry_two.push(a * 2)
arry_three.push(a * 3)
arry_five.push(a * 5)
arry.push(a)

a = arry_two[i += 1]
} else if(b < a && b < c ) {
arry_three.push(b * 3)
arry_five.push(b * 5)
arry.push(b)

b = arry_three[j += 1]
} else if (c < a && c < b) {
arry_five.push(c * 5)
arry.push(c)

c = arry_five[k += 1]
}
if(arry.length + 1 === n) break
}
//console.log(arry)
return arry[n - 2] //减去 0和 1
};

//console.log(nthUglyNumber(5))

现在来分析一下解决这个题目的思路:

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丑数就是只由2 3 5 乘出来的数字,我们定义三个数组,我们每次三个数组最小的数字,在第一组就分别乘以 2 3 5 放入 三个数组 
arry_two 2
arry_three 3
arry_five 5

arry_two 2 4
arry_three 3 6
arry_five 5 10
然后我们把最小的取出来
arry_two x 4
arry_three 3 6
arry_five 5 10
此时最小的是3,他位于第二个数组,我们给他 乘以3和5分别放入第二个和第三个数组
arry_two x 4
arry_three 3 6 9
arry_five 5 10 15
我们去出3,此时最小的是4,4 位于第一个数组,我们乘以2 3 5 后分别放入三个数组
arry_two x 4 8
arry_three x 6 9 12
arry_five 5 10 15 20
我们拿去4 后最小的是5,然后我们 给5 乘以5放入第三个数组
arry_two x x 8
arry_three x 6 9 12
arry_five 5 10 15 20 25

然后我们把每次拿出来的数字,放到一个数组,就是从2开始的丑数啦

在再放出我一开始的问题程序:

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var nthUglyNumber = function(n) {
//debugger
var ff = n
n = n + 150
var count = 4 //1也是丑数
var arry = []
var arry_temp = []
var arry_two = [2]
var i = 0
var arry_three = [3]
var j = 0
var arry_five = [5]
var k = 0

var a = arry_two[i]
var b = arry_three[i]
var c = arry_five[i]

if (ff < 5) {
return ff
}
while (1) {
if ( a < b && a < c) {
arry_two.push(a * 2)
count++
if(count === n) break

arry_three.push(a * 3)
count++
if(count === n) break

arry_five.push(a * 5)
count++
if(count === n) break

a = arry_two[i += 1]
} else if(b < a && b < c ) {
arry_three.push(b * 3)
count++
if(count === n) break

arry_five.push(b * 5)
count++
if(count === n) break

b = arry_three[j += 1]
} else if (c < a && c < b) {
arry_five.push(c * 5)
count++
if(count === n) break

c = arry_five[k += 1]
}

}
arry_temp = arry_two.concat(arry_three)
arry = arry_temp.concat(arry_five)
function sortNumber(a,b){
return a - b
}
arry.sort(sortNumber)
//console.log(arry)
return arry[n - 152] //多算了x个 减去 0和 1
};

这个程序的思路没有放在取出来的丑数上,而是放在了求出来的丑数上,我计算出目前算了多少个丑数,比如要10个我就计算10个,结果发现求出来的并没有按顺序(中间有空缺),比如算10个丑数,其实只计算到了15,计算第11个丑数的时候才把12计算出来,当时头脑一发热,好不是少计算了,那我多计算好了,所以我们多计算150个,结果在1 和 1690 的时候都对,在(1406,1430)之间不对,其余数字都对,真让人心累,为什么出现这样情况,现在头脑不清楚,有空好好进去瞅瞅,应该还是少计算的原因,不过比他大的数字都是好的,这个就很奇怪了,不过就算这个可以通过测试,比如我多计算500个或者600个。这个思路都是有问题的,不是好思路。

感觉不错的话给博主赞助一下呗