# 2460. Apply Operations to an Array

Easy
You are given a 0-indexed array `nums` of size `n` consisting of non-negative integers.
You need to apply `n - 1` operations to this array where, in the `ith` operation (0-indexed), you will apply the following on the `ith` element of `nums`:
• If `nums[i] == nums[i + 1]`, then multiply `nums[i]` by `2` and set `nums[i + 1]` to `0`. Otherwise, you skip this operation.
After performing all the operations, shift all the `0`'s to the end of the array.
• For example, the array `[1,0,2,0,0,1]` after shifting all its `0`'s to the end, is `[1,2,1,0,0,0]`.
Return the resulting array.
Note that the operations are applied sequentially, not all at once.
Example 1:
Input: nums = [1,2,2,1,1,0]
Output:
[1,4,2,0,0,0]
Explanation:
We do the following operations:
- i = 0: nums and nums are not equal, so we skip this operation.
- i = 1: nums and nums are equal, we multiply nums by 2 and change nums to 0. The array becomes [1,4,0,1,1,0].
- i = 2: nums and nums are not equal, so we skip this operation.
- i = 3: nums and nums are equal, we multiply nums by 2 and change nums to 0. The array becomes [1,4,0,2,0,0].
- i = 4: nums and nums are equal, we multiply nums by 2 and change nums 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].
Example 2:
Input: nums = [0,1]
Output:
[1,0]
Explanation:
No operation can be applied, we just shift the 0 to the end.
Constraints:
• `2 <= nums.length <= 2000`
• `0 <= nums[i] <= 1000`

### 解題

Runtime: 7 ms, faster than 100%
Memory Usage: 3.2 MB, less than 100%
func applyOperations(nums []int) []int {
length := len(nums)
res := make([]int, 0)
for i:=1; i<length; i++ {
if nums[i] == nums[i-1] {
nums[i-1] *= 2
nums[i] = 0
}
}
for _, num := range nums {
if num != 0 { res = append(res, num) }
}
for _, num := range nums {
if num == 0 { res = append(res, num) }
}
return res
}