func minimumNumbers(num int, k int) int { n if num == 0 { n return 0n }n if k == 0 { n if num%10 == 0 { n return 1n }n return -1n }n if num%2 == 1 && k%2 == 0 { n return -1n }n for i := 1; i <= num; i++ { n sum := i * kn left := num - sumn if left%10 == 0 { n return in }n if left < 0 { n return -1n }n }n return -1n}
2 、枚举;时间复杂度O(1),空间复杂度O(1)
func minimumNumbers(num int, k int) int { n if num == 0 { n return 0n }n for i := 1; i <= 10; i++ { n // 只考虑个位数n if i*k <= num && (i*k)%10 == num%10 { n return in }n }n return -1n}