[Leetcode] Ugly Number II, Solution

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.
[Thoughts]这就是多链表Merge Sort的一个扩展题。对于任意一个ugly number - K, 2*K, 3*K, 和5*K都是ugly number，所以说新的ugly number都是从已有的ugly number上，通过与{2,3,5}相乘而产生的。如果Ugly Number: 1, 2, 3, 4, 5, 6, 8, 10, ..............那么 1*2 2*2 3*2 4*2 5*2 6*2 8*2 10*2 .............. *2 1*3 2*3 3*3 4*3 5*3 6*3 8*3 10*3 .............. *3 1*5 2*5 3*5 4*5 5*5 6*5 8*5 10*5 .............. *5都是ugly number。只要不断把新产生的ugly number通过merge sort添加到原有的ugly number数组中就可以了，直到找到第N个。

[code]1:  class Solution {
2:  public:
3:    int nthUglyNumber(int n) {
4:      vector<int> uglys(1, 1);
5:      int p2 = 0, p3 = 0, p5 = 0;
6:      while (uglys.size() < n) {
7:        int ugly2 = uglys[p2] * 2, ugly3 = uglys[p3] * 3, ugly5 = uglys[p5] * 5;
8:        int min_v = min(ugly2, min(ugly3, ugly5));
9:        if (min_v == ugly2) ++p2;
10:        if (min_v == ugly3) ++p3;
11:        if (min_v == ugly5) ++p5;
12:        if(min_v != uglys.back()) {
13:          // skip duplicate
14:          uglys.push_back(min_v);
15:        }
16:      }
17:      return uglys[n-1];
18:    }
19:  };

考虑到通用性，可以扩展如下，可以支持任意长度的因子数组factors。

1:  class Solution {
2:  public:
3:    int nthUglyNumber(int n) {
4:      vector<int> factors{ 2, 3, 5};
5:      return nthUglyNumberGeneral(n, factors);
6:    }
7:    int nthUglyNumberGeneral(int n, vector<int>& factors) {
8:      vector<int> uglys(1,1);
9:      vector<int> indexes(factors.size(), 0);
10:      while(uglys.size() < n) {
11:        int min_v = INT_MAX;
12:        int min_index = 0;
13:        for(int k =0; k< factors.size(); k++) {
14:          int temp = uglys[indexes[k]] * factors[k];
15:          if(temp < min_v) {
16:            min_v = temp;
17:            min_index = k;
18:          }
19:        }
20:        indexes[min_index]++;
21:        // need to avoid duplicate ugly number
22:        if(uglys[uglys.size()-1] != min_v) {
23:          uglys.push_back(min_v);
24:         }
25:      }
26:      return uglys[n-1];
27:    }
28:  };

从空间的优化来说，没有必要用一个uglys的数组保存所有的ugly number，尤其是当n是个非常大的数字。对于indexes指针扫过的ugly number，都可以丢掉了。不过，懒得写了。