LeetCode: 313. Super Ugly Number

LeetCode: 313. Super Ugly Number

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.

Example:

Input: n = 12, primes = [2,7,13,19]
Output: 32 
Explanation: [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 is a super ugly number for any given primes.
• The given numbers in primes are in ascending order.
• 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.
• The nth super ugly number is guaranteed to fit in a 32-bit signed integer.

解题思路

1. 动态规划的思路：ugly number = ugly number * primes
2. 记录每个 ​​primes​​ 下一次应该用的 ugly number。每次选择最小的 ugly number * primes 作为新的 ugly number

AC 代码

func min(lhv, rhv int) int {
    if lhv < rhv {
        return lhv
    } else {
        return rhv
    }
}

func nthSuperUglyNumber(n int, primes []int) int {
    var uglynum []int // 记录前 n 个 uglynum
    var records []int // records[i] 记录下一个需要乘以 primes[i] 的 uglynum 的下标
    
    //初始化:
    // 第一个 ugly number 是 1
    uglynum = append(uglynum, 1)
    // 一开始所有的 primes 下一个要乘的数都是下标为 0 的ugly number 
    for i := 0; i < len(primes); i++ {
        records = append(records, 0)
    }
    
    for i := 1; i < n; i++ {
        for j := 0; j < len(primes); j++ {
            cur := primes[j]*uglynum[records[j]]
            if cur == uglynum[i-1] { // 上一个循环用过的 primes
                records[j]++
                cur = primes[j]*uglynum[records[j]]
            }
            if j == 0 { // 下一个 ugly number
                uglynum = append(uglynum, cur)
            } else {
                uglynum[i] = min(uglynum[i], cur)
            }
        }
    }
    
    return uglynum[n-1]
    
}