add 2169

Author: clingoram

javascript/LeetCode/Array/2169.js

@@ -0,0 +1,42 @@
+/**
+ * 2169. Count Operations to Obtain Zero
+ * 
+ * 一次操作中，若nums1 > nums2，則nums1 = nums1 - nums2
+ * nums1 < nums2， nums2 = nums1 - nums2
+ * 計算要幾次才能使得nums1 = 0 or num2 = 0
+ * 
+ * @param {number} num1
+ * @param {number} num2
+ * @return {number}
+ */
+var countOperations = function(num1, num2) {
+  // solution 1.
+  // let ans = 0;
+  // while(num1 !== 0 && num2 !== 0){
+  //     ans += Math.floor(num1 / num2);
+  //     num1 %= num2;
+  //     [num1,num2] = [num2,num1];
+  // }
+  // return ans;
+
+  // solution 2.
+  let ans = 0;
+  while(num1 !== 0 && num2 !== 0){
+      ans += Math.floor(num1 / num2);
+      num1 %= num2;
+      // swap nums1 and nums2
+      let temp = num1;
+      num1 = num2;
+      num2 = temp;
+  }
+  return ans;
+};
+let 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.
+console.log(countOperations(num1,num2))

javascript/index.js

@@ -1079,35 +1079,3 @@ var findXSum = function(nums, k, x) {
 // console.log(findXSum(nums,k,x));
 
 
-/**
- * 2169. Count Operations to Obtain Zero
- * 
- * 一次操作中，若nums1 >= nums2，則nums1 = nums1 - nums2，否則nums2 = nums1 - nums2
- * 計算要幾次才能使得nums1 = 0 or num2 = 0
- * 
- * @param {number} num1
- * @param {number} num2
- * @return {number}
- */
-var countOperations = function(num1, num2) {
-    let ans = 0;
-    while(num1 !== 0 || num2 !== 0){
-      if(num1 > num2){
-        num1 = num1 - num2;
-        ans++;
-      }else{
-        num2 = num1 - num2;
-        ans++;
-      }
-    }
-    return ans;
-};
-let 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.
-console.log(countOperations(num1,num2))