add JS one problem solution

Author: clingoram

javascript/LeetCode/Array/2460.js

@@ -0,0 +1,39 @@
+/**
+ * 2460. Apply Operations to an Array
+ * 
+ * 操作n - 1次
+ * 若nums[i] === nums[i +1]，則將nums[i] * 2，將nums[i+1]改成0
+ * 完成所有操作，將所有的0移到陣列最後面
+ * 回傳陣列
+ * @param {number[]} nums
+ * @return {number[]}
+ */
+var applyOperations = function(nums) {
+    let j = 0;
+    for(let i = 0;i < nums.length - 1;++i) {
+      if(nums[i] === nums[i+1]){
+        nums[i] *= 2;
+        nums[i + 1] = 0;
+      }
+    }
+    for(let i = 0;i < nums.length;++i) {
+      if(nums[i] !== 0){
+        nums[j] = nums[i];
+        j++;
+      }
+    }
+    while (j < nums.length) {
+      nums[j++] = 0;
+    }
+    return nums;
+};
+let nums = [1,2,2,1,1,0]
+// Output: [1,4,2,0,0,0]
+// Explanation: We do the following operations:
+// i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
+// i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
+// i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
+// i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
+// i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
+// After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].
+console.log(applyOperations(nums));

python/oop/Tortoise.py

@@ -13,7 +13,7 @@ def __init__(self, name, age, food,speciceType:str):
         self.speciceType = speciceType
 
     def eat(self):
-        cant = ["草酸高食物","糖份高，例如水果"]
+        cant = ["草酸高食物","糖份高，例如水果","人類的食物"]
 
         print(f"{self.food}，但有些不適合吃")