Leetcode 2169. Count Operations to Obtain Zero

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 = 3Output: 3Explanation: - 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 = 10Output: 1Explanation: - 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 <= 105

• Easy 题目，但是没想到能超过100%；

• 大的减小的，就一步到位，直接减小数的倍数，这样生效的肯定是较小的数，

• 然后while循环即可；

Runtime: 0 ms, faster than 100.00% of Java online submissions for Count Operations to Obtain Zero.

Memory Usage: 39.7 MB, less than 30.54% of Java online submissions for Count Operations to Obtain Zero.