【LeetCode】0026. Remove Duplicates from Sorted Array

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Given an integer array nums sorted in non-decreasing order, remove the duplicates in-place such that each unique element appears only once. 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:

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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.


Example 1:

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


Example 2:

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Input: nums = [0,0,1,1,1,2,2,3,3,4]
Output: 5, nums = [0,1,2,3,4,_,_,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums being 0, 1, 2, 3, and 4 respectively.
It does not matter what you leave beyond the returned k (hence they are underscores).


Constraints:

  • 1 <= nums.length <= 3 * 104
  • -100 <= nums[i] <= 100
  • nums is sorted in non-decreasing order.


Related Topics: Array Two Pointers

解題邏輯與實作

這題是要我們從 Array 中去除重複項。這邊我第一個想法是使用 for 迴圈，我另外用了兩個指標 compar_idx 與 replace_idx 以記錄目前的位置，並用 for 迴圈尋訪 Array 內容。一旦目前被尋訪到的值與比較值（compar_idx 所指的值）相同，就將該值指定給取代值（replace_idx 所指的值），完成後就將 compar_idx 移到目前位置且 replace_idx 也向前移動一步：

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def removeDuplicates(nums):
    if not nums:
        return 0
    
    length = len(nums)
    if length <= 1:
        return length

    compar_idx = 0
    replace_idx = 1
  
    for i in range(1,length): 
        if nums[compar_idx] != nums[i]:
            nums[replace_idx] = nums[i]
            compar_idx = i
            replace_idx += 1
    return replace_idx

出來效果還行 83 ms，93.30%。

其他連結

  1. 【LeetCode】0000. 解題目錄

更新紀錄

最後更新日期：2023-02-15
  • 2023-02-15 發布
  • 2023-02-02 完稿
  • 2023-02-02 起稿