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:

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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:

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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

题目又臭又长，简而言之就是先给数组操作一番，然后把所有非零元素移到前面，0移到后面。

前面的操作没啥好说的。后面移0们依旧是双指针。

根据27的经验，用了俩指针，一个slow从0开始，一个fast从1开始。如果slow已经是非零了就++，如果slow是0但是fast不是0就交换且slow++。不管咋样都要fast++遍历。其实自己也想的不是那么清楚，试了几次以后才过。

后来看了答案，比我自己写的简单。一个nonZero表示在这个元素之前的全是非0（注意是之前，而不是它和它之前），另一个i从0开始遍历。直接考虑在遍历的那个i，如果它不是0，那就把nonZero设成它，此时当然需要nonZero++。因为i一直在遍历所以不管怎样都要++。最后，i遍历完了以后因为我们没有把后面的重新assign成0，所以需要把nonZero及其后面的都设成0。

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

        int nonZero = 0;  // elements before this are nonZero
        i = 0;

        while (i < nums.length) {
            if (nums[i] != 0) {
                nums[nonZero] = nums[i];
                nonZero++;
            }
            i++;
        }

        while (nonZero < nums.length) {
            nums[nonZero] = 0;
            nonZero++;
        }

        // original:
        // int slow = 0;
        // int fast = 1;
        // while (fast < nums.length) {
        //     if (nums[slow] != 0) {
        //         slow+;
        //     }
        //     if (nums[slow] == 0 && nums[fast] != 0) {
        //         nums[slow] = nums[fast];
        //         nums[fast] = 0;
        //         slow++;
        //     }
        //     fast+;
        // }
        return nums;
    }
}