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[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].

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
}