## Description
>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.

## Constraints:
> 2 <= nums.length <= 2000
0 <= nums[i] <= 1000

## Example1
Input: 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].

## Example2
Input: nums = [0,1]
Output: [1,0]
Explanation: No operation can be applied, we just shift the 0 to the end.

## Solution
解題思路都寫在描述當中，nums的第i個和第i+1個做比較，不相等的話就不做任何動作，相等的話nums[i]的值乘二，nums[i+1]的值歸0，遍歷完陣列之後，最後要再把所有0的值shift到陣列的尾端。

* Time complexity: O(n)
* Space complexity: O(n)

## Code
### C++

class Solution {
public:
    vector<int> applyOperations(vector<int>& nums) {
        
        int n = nums.size();
        
        for(int i = 0; i < n-1 ; i++) {
            if(nums[i] == nums[i+1])
            {
                nums[i] = nums[i] * 2;
                nums[i+1] = 0;
            }    
        }
        int idx = 0;
        for(int i = 0 ; i < n ; i++) {
            if(nums[i] != 0) {
                nums[idx++] = nums[i];        
            }
        }
        for(int i = idx ; i < n ; i++) {
            nums[i] = 0;            
        }
        return nums;
    }
};