264. Ugly Number II

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题目：

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

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note that 1 is typically treated as an ugly number.

Hint:

1. The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3. The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
4. Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

答案：

找出第n个丑陋数，排除依次判断是否为丑陋数的做法，根据题目的提示，可以以2,3,5为基准数，分为三个数列，每次从三个数列中取一个数字，选出最小的那个。

实现代码如下：

 1 class Solution {
 2 public:
 3     int nthUglyNumber(int n) {
 4         int arr[n];
 5         arr[0]=1;
 6         int num2=2,num3=3,num5=5;
 7         int count2=0,count3=0,count5=0;
 8         for(int i=1;i<n;i++){
 9             arr[i]=getMin(num2,num3,num5);
10             if(arr[i]==num2){
11                 num2=2*arr[++count2];
12             }
13             if(arr[i]==num3){
14                 num3=3*arr[++count3];
15             }
16             if(arr[i]==num5){
17                 num5=5*arr[++count5];
18             }
19         }
20         return arr[n-1];
21     }
22     int getMin(int a,int b,int c){
23         int minNum=(a<=b?a:b);
24         minNum=(minNum<=c?minNum:c);
25         return minNum;
26     }
27 };

注意：第11,14,17行的加号是在count前面的，先加再取数

264. Ugly Number II

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原文地址：http://www.cnblogs.com/Reindeer/p/5727361.html