题目:
Write a program to find the nth super ugly number.
Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes
of size k
. For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32]
is the sequence of the first 12 super ugly numbers given primes
= [2, 7, 13, 19]
of size 4.
Note:
(1) 1
is a super ugly number for any given primes
.
(2) The given numbers in primes
are in ascending order.
(3) 0 < k
≤ 100, 0 < n
≤ 106, 0 < primes[i]
< 1000.
链接: http://leetcode.com/problemset/algorithms/
题解:
超级丑数。跟丑数II很类似,只不过这次primes从2, 3, 5变成了一个size k的list。方法应该有几种,一种是维护一个size K的min-oriented heap,heap里是k个queue,和Ugly Number II的方法一样,取最小的那一个,然后更新其他Queue里的元素,n--,最后n = 1时循环结束。 另外一种是类似dynamic programming的方法,主要参考了larrywant2014大神的代码。维护一个index数组,维护一个dp数组。每次遍历更新的转移方程非常巧妙,min = dp[[index[j]]] * primes[j]。之后再便利一次primes来update每个数在index[]中的使用次数。
Time Complexity - O(nk), Space Complexity - O(n + k).
public class Solution {
public int nthSuperUglyNumber(int n, int[] primes) {
if(n < 1) {
return 0;
}
int len = primes.length;
int[] index = new int[len];
int[] dp = new int[n];
dp[0] = 1; for(int i = 1; i < n; i++) {
int min = Integer.MAX_VALUE;
for(int j = 0; j < len; j++) { // try update with all primes
min = Math.min(dp[index[j]] * primes[j], min);
}
dp[i] = min; // find dp[i]
for(int j = 0; j < len; j++) { //if prices[j] is used, increase the index
if(dp[i] % primes[j] == 0) {
index[j]++;
}
}
}
return dp[n - 1];
}
}
题外话:
休假结束,又开始上班啦。不能象前几天一样愉快地刷题了...
Reference:
https://leetcode.com/discuss/72878/7-line-consice-o-kn-c-solution
https://leetcode.com/discuss/72835/108ms-easy-to-understand-java-solution
https://leetcode.com/discuss/74433/simple-addiction-logic-reduce-the-runtime-28ms-and-beats-77%25
http://bookshadow.com/weblog/2015/12/03/leetcode-super-ugly-number/
http://www.cnblogs.com/Liok3187/p/5016076.html
http://www.hrwhisper.me/leetcode-super-ugly-number/
http://my.oschina.net/Tsybius2014/blog/547766?p=1