题目 #

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

思路1 #

分析 #

• 照着做，复用原数组，搞一个非0的索引

代码 #

 1func applyOperations(nums []int) []int {
 2	n := len(nums)
 3	j := 0 // j是非0的索引
 4	for i := 0; i < n; i++ {
 5		if i == n-1 {
 6			nums[j] = nums[i]
 7			j++
 8			continue
 9		}
10		if nums[i] > 0 {
11			if nums[i] == nums[i+1] {
12				nums[j] = nums[i] * 2
13				nums[i+1] = 0
14				i++
15			} else {
16				nums[j] = nums[i]
17			}
18			j++
19		}
20	}
21	for ; j < n; j++ {
22		nums[j] = 0
23	}
24	return nums
25}