題目:
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.
鏈接: http://leetcode.com/problemset/algorithms/
題解:
超級丑數。跟丑數II很類似,只不過這次primes從2, 3, 5變成了一個size k的list。方法應該有幾種,一種是維護一個size K的min-oriented heap,heap里是k個queue,和Ugly Number II的方法一樣,取最小的那一個,然后更新其他Queue里的元素,n--,最后n = 1時循環結束。 另外一種是類似dynamic programming的方法,主要參考了larrywant2014大神的代碼。維護一個index數組,維護一個dp數組。每次遍歷更新的轉移方程非常巧妙,min = dp[[index[j]]] * primes[j]。之后再便利一次primes來update每個數在index[]中的使用次數。
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]; } }
題外話:
休假結束,又開始上班啦。不能象前幾天一樣愉快地刷題了...
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