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 題目,但是沒想到能超過100%;
大的減小的,就一步到位,直接減小數(shù)的倍數(shù),這樣生效的肯定是較小的數(shù),
然后while循環(huán)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.