Lintcode: Kth Prime Number (Original Name: Ugly Number)
Ugly number is a number that only have factors3
,5
and7.
Design an algorithm to find the kth number such that the only prime factors are 3, 5, and 7. The eligible numbers are like 3, 5, 7, 9, 15 ... Example
If k=4, return 9. Challenge
O(n log n) or O(n) time
这是丑数问题(Ugly Number), 思路参看:
http://www.cppblog.com/zenliang/articles/131094.html
http://www.geeksforgeeks.org/ugly-numbers/
分析:假设数组ugly[N]中存放不断产生的丑数,初始只有一个丑数ugly[0]=1,由此出发,下一个丑数由因子2,3,5竞争产生,得到ugly[0]*2, ugly[0]*3, ugly[0]*5, 显然最小的那个数是新的丑数,所以第2个丑数为ugly[1]=2,开始新一轮的竞争,由于上一轮竞争中,因子2获胜,这时因子2应该乘以ugly[1]才显得公平,得到ugly[1]*2,ugly[0]*3,ugly[0]*5, 因子3获胜,ugly[2]=3,同理,下次竞争时因子3应该乘以ugly[1],即:ugly[1]*2, ugly[1]*3, ugly[0]*5, 因子5获胜,得到ugly[3]=5,重复这个过程,直到第n个丑数产生。总之:每次竞争中有一个(也可能是两个)因子胜出,下一次竞争中 胜出的因子就应该加大惩罚!
注意这里不可以使用if/else 循环,因为有可能多于一个指针的结果是相等的:例如p3->5, p5->3, 他们的结果相等,这是两个指针都要+1
class Solution {
/**
* @param k: The number k.
* @return: The kth prime number as description.
*/
public long kthPrimeNumber(int k) {
// write your code here
long[] res = new long[k+1];
res[0] = 1;
int k3 = 0, k5 = 0, k7 = 0;
for (int i=1; i<=k; i++) {
res[i] = Math.min(Math.min(res[k3]*3, res[k5]*5), res[k7]*7);
if (res[i]/res[k3] == 3) k3++;
if (res[i]/res[k5] == 5) k5++;
if (res[i]/res[k7] == 7) k7++;
}
return res[k];
}
};
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