LeetCode 6004. Get operands of 0

6004. obtain 0 The number of operations

Here are two for you non-negative Integers num1 and num2 .

Each step operation in , If num1 >= num2 , You have to use num1 reduce num2 ; otherwise , You have to use num2 reduce num1 .

• for example ,num1 = 5 And num2 = 4 , Should use the num1 reduce num2 , therefore , obtain num1 = 1 and num2 = 4 . However , If num1 = 4 And num2 = 5 , After one step operation , obtain num1 = 4 and num2 = 1 .

Return to make num1 = 0 or num2 = 0 Of Operands .

Example 1：

 Input ：num1 = 2, num2 = 3
 Output ：3
 explain ：
-  operation  1 ：num1 = 2 ,num2 = 3 . because  num1 < num2 ,num2  reduce  num1  obtain  num1 = 2 ,num2 = 3 - 2 = 1 .
-  operation  2 ：num1 = 2 ,num2 = 1 . because  num1 > num2 ,num1  reduce  num2 .
-  operation  3 ：num1 = 1 ,num2 = 1 . because  num1 == num2 ,num1  reduce  num2 .
 here  num1 = 0 ,num2 = 1 . because  num1 == 0 , No more action is required .
 So the total operand is  3 .

Example 2：

 Input ：num1 = 10, num2 = 10
 Output ：1
 explain ：
-  operation  1 ：num1 = 10 ,num2 = 10 . because  num1 == num2 ,num1  reduce  num2  obtain  num1 = 10 - 10 = 0 .
 here  num1 = 0 ,num2 = 10 . because  num1 == 0 , No more action is required .
 So the total operand is  1 .

Tips ：

• 0 <= num1, num2 <= 105

Two 、 Method 1

simulation , That is, division by turns

class Solution {
    
    public int countOperations(int num1, int num2) {
    
        int res = 0;
        while (num1 != 0 && num2 != 0) {
    
            if (num1 >= num2) {
    
                num1 -= num2;
            } else {
    
                num2 -= num1;
            }
            res++;
        } 
        return res;
    }
}

Complexity analysis

• Time complexity ：O(n).

• Spatial complexity ：O(1).