題目鏈接
tag:
- Medium；
question：
??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.
Hint:
- 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.
- An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
- 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.
- Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).
思路：
??本題是之前那道Ugly Number 丑數(shù)的延伸憔杨，讓我們找第n個丑陋數(shù)脱柱，題目中給了很多提示坞生，基本上相當(dāng)于告訴我們解法了岔留，根據(jù)提示中的信息吨拗，我們知道丑陋數(shù)序列可以拆分為下面3個子列表：
(1) 1x2, 2x2, 2x2, 3x2, 3x2, 4x2, 5x2...
(2) 1x3, 1x3, 2x3, 2x3, 2x3, 3x3, 3x3...
(3) 1x5, 1x5, 1x5, 1x5, 2x5, 2x5, 2x5...
仔細(xì)觀察上述三個列表庵佣，我們可以發(fā)現(xiàn)每個子列表都是一個丑陋數(shù)分別乘以2,3,5歉胶，而要求的丑陋數(shù)就是從已經(jīng)生成的序列中取出來的，我們每次都從三個列表中取出當(dāng)前最小的那個加入序列巴粪，代碼如下：
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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