2169. Count Operations to Obtain Zero

题目

You are given two non-negative integers num1 and num2.

In one operation, if num1 >= num2, you must subtract num2 from num1, otherwise subtract num1 from num2.

For example, if num1 = 5 and num2 = 4, subtract num2 from num1, thus obtaining num1 = 1 and num2 = 4. However, if num1 = 4 and num2 = 5, after one operation, num1 = 4 and num2 = 1.

Return the number of operations required to make either num1 = 0 or num2 = 0.

Example 1:

Input: num1 = 2, num2 = 3
Output: 3
Explanation:
- Operation 1: num1 = 2, num2 = 3. Since num1 < num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1.
- Operation 2: num1 = 2, num2 = 1. Since num1 > num2, we subtract num2 from num1.
- Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1.
Now num1 = 0 and num2 = 1. Since num1 == 0, we do not need to perform any further operations.
So the total number of operations required is 3.

Example 2:

Input: num1 = 10, num2 = 10
Output: 1
Explanation:
- Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0.
Now num1 = 0 and num2 = 10. Since num1 == 0, we are done.
So the total number of operations required is 1.

Constraints:

0 <= num1, num2 <= 10^5

题目大意

给你两个非负整数 num1 和 num2 。每一步操作中，如果 num1 >= num2 ，你必须用 num1 减 num2 ；否则，你必须用 num2 减 num1 。

例如，num1 = 5 且 num2 = 4 ，应该用 num1 减 num2 ，因此，得到 num1 = 1 和 num2 = 4 。然而，如果 num1 = 4且 num2 = 5 ，一步操作后，得到 num1 = 4 和 num2 = 1 。

返回使 num1 = 0 或 num2 = 0 的操作数。

解题思路

简单题，按照题意模拟，每次两个数字相减，便累加操作次数。当某个数字变为 0 时，输出操作次数。

代码

package leetcode

func countOperations(num1 int, num2 int) int {
	res := 0
	for num1 != 0 && num2 != 0 {
		if num1 >= num2 {
			num1 -= num2
		} else {
			num2 -= num1
		}
		res++
	}
	return res
}

Uh oh!