Leetcode 80. Remove Duplicates from Sorted Array II (Medium)

Leetcode Top Interview 150 的其中一題

題目描述

Given an integer array nums sorted in non-decreasing order, remove some duplicates in-place such that each unique element appears at most twice. The relative order of the elements should be kept the same.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the first part of the array nums. More formally, if there are k elements after removing the duplicates, then the first k elements of nums should hold the final result. It does not matter what you leave beyond the first k elements.

Return k after placing the final result in the first k slots of nums.

Do not allocate extra space for another array. You must do this by modifying the input array in-place with O(1) extra memory.

Custom Judge:

The judge will test your solution with the following code:

int[] nums = [...]; // Input array
int[] expectedNums = [...]; // The expected answer with correct length

int k = removeDuplicates(nums); // Calls your implementation

assert k == expectedNums.length;
for (int i = 0; i < k; i++) {
    assert nums[i] == expectedNums[i];
}

If all assertions pass, then your solution will be accepted.

範例

Example 1:

Input: nums = [1,1,1,2,2,3]
Output: 5, nums = [1,1,2,2,3,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 1, 1, 2, 2 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Example 2:

Input: nums = [0,0,1,1,1,1,2,3,3]
Output: 7, nums = [0,0,1,1,2,3,3,_,_]
Explanation: Your function should return k = 7, with the first seven elements of nums being 0, 0, 1, 1, 2, 3 and 3 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).

Constraints:

• 1 <= nums.length <= 3 * 104
• -104 <= nums[i] <= 104
• nums is sorted in non-decreasing order.

解法

這一題和上一題 26. 類似，只是我們要讓每個數字只能出現兩次，要在原本的陣列內實作且只能有 O(1) 的 extra memory。 方法：檢查每個數字現在是否是出現第三次以上，是的話改成 100001，這樣排序後就會在最後面，不在列入檢查的陣列前面

func removeDuplicates(nums []int) int {
    count := 1
    now := nums[0]
    ans := len(nums)

    for i:=1; i<len(nums); i++ {
        if nums[i] == now {
            count++
        } else {
            count = 1
            now= nums[i]
        }

        if count > 2 {
            nums[i] = 100001
            ans--
        }
    }

    sort.Ints(nums)

    return ans
}