問(wèn)題描述：

A?Hamming number?is a positive integer of the form 2ⁱ3^j5^k, for some non-negative integers?i,?j, and?k.

Write a function that computes the?nth smallest Hamming number.

Specifically:

• The first smallest Hamming number is 1 = 2⁰3⁰5⁰
• The second smallest Hamming number is 2 = 2¹3⁰5⁰
• The third smallest Hamming number is 3 = 2⁰3¹5⁰
• The fourth smallest Hamming number is 4 = 2²3⁰5⁰
• The fifth smallest Hamming number is 5 = 2⁰3⁰5¹

The 20 smallest Hamming numbers are given in example test fixture.

Your code should be able to compute all of the smallest 5,000 (Clojure: 2000) Hamming numbers without timing out.

我的思路：

本題自己是沒有任何思路的，只是知道漢明數(shù)肯定是2或3或5的倍數(shù)，但是無(wú)從下手。后來(lái)看別人的答案，主要思路也是如此。

下一個(gè)漢明數(shù)為已存在漢明數(shù)的2x，3x，5x的倍數(shù)。

若i2是我們沒有用過(guò)的漢明數(shù)的指數(shù)的話，就乘以2；

若i3是我們沒有用過(guò)的漢明數(shù)的指數(shù)的話，就乘以3；

若i5是我們沒有用過(guò)的漢明數(shù)的指數(shù)的話，就乘以5。

我的答案：無(wú)，o(╥﹏╥)o

優(yōu)秀答案：

（1）

function hamming (n) {var seq = [1];var i2 = 0, i3 = 0, i5 = 0;for (var i = 1; i < n; i++) {var x = Math.min(2 * seq[i2], 3 * seq[i3], 5 * seq[i5]);seq.push(x);if (2 * seq[i2] <= x) i2++; //<= 可換成 ==if (3 * seq[i3] <= x) i3++;if (5 * seq[i5] <= x) i5++;}return seq[n-1]; }

?

哈哈哈！

轉(zhuǎn)載于:https://www.cnblogs.com/hiluna/p/8916419.html

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