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

新题中没有提示,老版的有。

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).

写一个程序找到第n个丑陋数,根据提示,丑陋数序列可以拆分为下面的子列表:

(1) 1x2,  2x2, 2x2, 3x2, 3x24x2, 5x2...
(2) 1x3,  1x3, 2x3, 2x3, 2x3, 3x3, 3x3...
(3) 1x5,  1x5, 1x5, 1x5, 2x5, 2x5, 2x5...

每个子列表都是一个丑陋数分别乘以2,3,5,而要求的丑陋数就是从已经生成的序列中取出来的,每次都从三个列表中取出当前最小的那个加入序列。

Java:

public class Solution {
public int nthUglyNumber(int n) {
int[] ugly = new int[n];
ugly[0] = 1;
int index2 = 0, index3 = 0, index5 = 0;
int factor2 = 2, factor3 = 3, factor5 = 5;
for(int i=1;i<n;i++){
int min = Math.min(Math.min(factor2,factor3),factor5);
ugly[i] = min;
if(factor2 == min)
factor2 = 2*ugly[++index2];
if(factor3 == min)
factor3 = 3*ugly[++index3];
if(factor5 == min)
factor5 = 5*ugly[++index5];
}
return ugly[n-1];
}
}

Python:

def nthUglyNumber(self, n):
ugly = [1]
i2, i3, i5 = 0, 0, 0
while n > 1:
u2, u3, u5 = 2 * ugly[i2], 3 * ugly[i3], 5 * ugly[i5]
umin = min((u2, u3, u5))
if umin == u2:
i2 += 1
if umin == u3:
i3 += 1
if umin == u5:
i5 += 1
ugly.append(umin)
n -= 1
return ugly[-1] 

C++:

class Solution {
public:
int nthUglyNumber(int n) {
if(n <= 0) return false; // get rid of corner cases
if(n == 1) return true; // base case
int t2 = 0, t3 = 0, t5 = 0; //pointers for 2, 3, 5
vector<int> k(n);
k[0] = 1;
for(int i = 1; i < n ; i ++)
{
k[i] = min(k[t2]*2,min(k[t3]*3,k[t5]*5));
if(k[i] == k[t2]*2) t2++;
if(k[i] == k[t3]*3) t3++;
if(k[i] == k[t5]*5) t5++;
}
return k[n-1];
}
};  

C++:

class Solution {
public:
int nthUglyNumber(int n) {
vector<int> res(1, 1);
int i2 = 0, i3 = 0, i5 = 0;
while (res.size() < n) {
int m2 = res[i2] * 2, m3 = res[i3] * 3, m5 = res[i5] * 5;
int mn = min(m2, min(m3, m5));
if (mn == m2) ++i2;
if (mn == m3) ++i3;
if (mn == m5) ++i5;
res.push_back(mn);
}
return res.back();
}
};

  

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[LeetCode] 263. Ugly Number 丑陋数

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