[LeetCode] 2460. Apply Operations to an Array

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

对数组执行操作。

给你一个下标从 0 开始的数组 nums ，数组大小为 n ，且由 非负 整数组成。
你需要对数组执行 n - 1 步操作，其中第 i 步操作（从 0 开始计数）要求对 nums 中第 i 个元素执行下述指令：
如果 nums[i] == nums[i + 1] ，则 nums[i] 的值变成原来的 2 倍，nums[i + 1] 的值变成 0 。否则，跳过这步操作。
在执行完 全部 操作后，将所有 0 移动 到数组的 末尾 。
例如，数组 [1,0,2,0,0,1] 将所有 0 移动到末尾后变为 [1,2,1,0,0,0] 。
返回结果数组。
注意 操作应当 依次有序 执行，而不是一次性全部执行。
来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/apply-operations-to-an-array
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思路

思路是追击型双指针，如果没有做过283题，建议先做一下，本题跟283题几乎一样。

复杂度

时间O(n)
空间O(n)

代码

class Solution {
    public int[] applyOperations(int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n - 1; i++) {
            if (nums[i] == nums[i + 1]) {
                nums[i] = nums[i] * 2;
                nums[i + 1] = 0;
            }
        }

        int cur = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] != 0) {
                nums[cur] = nums[i];
                cur++;
            }
        }
        while (cur < n) {
            nums[cur] = 0;
            cur++;
        }
        return nums;
    }
}

相关题目

283. Move Zeroes
2460. Apply Operations to an Array