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πŸ’‘ LeetCode 26 - Remove Duplicates from Sorted Array

πŸ’‘ LeetCode 26 - Remove Duplicates from Sorted Array

문제

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. Then return the number of unique elements in nums.

Consider the number of unique elements of nums to be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the unique elements in the order they were present in nums initially. The remaining elements of nums are not important as well as the size of nums. Return k. Custom Judge:

The judge will test your solution with the following code: ```java 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]; }

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


### μž…μΆœλ ₯ 예제
#### βœ… 예제 1
```bash
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).

βœ… 예제 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).

μ œμ•½μ‘°κ±΄

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

μž‘μ„± μ½”λ“œ

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class Solution {
	public int removeDuplicates(int[] nums) {
		// 1. λ³€μˆ˜ μ„ μ–Έ 및 μ΄ˆκΈ°ν™”
		Map<Integer, Integer> numMap = new HashMap<>();
		int idx = 0;
		
		// 2. λ°°μ—΄ μˆœνšŒν•˜λ©° 쀑볡 μ•„λ‹Œ κ°’μœΌλ‘œ μž¬ν• λ‹Ή
		for (int num : nums) {
			if (!numMap.containsKey(num)) {
				numMap.put(num, 1);
				nums[idx++] = num;
			}
		}
		
		// 3. λ°˜ν™˜
		return idx;
	}
}

  • HashMap을 λΆˆν•„μš”ν•˜κ²Œ μ‚¬μš©ν•œ λŠλ‚Œμ΄ λ“€μ–΄μ„œ μ½”λ“œλ₯Ό κ°œμ„  해보렀고 ν•œλ‹€.

κ°œμ„  μ½”λ“œ

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class Solution {
	public int removeDuplicates(int[] nums) {
		// 1. λ³€μˆ˜ μ„ μ–Έ 및 μ΄ˆκΈ°ν™”
		int idx = 0;
		int beforeNum = 999;
		
		// 2. 이전 κ°’κ³Ό ν˜„μž¬ 값이 λ‹€λ₯Ό 경우 λ°°μ—΄ κ°±μ‹ 
		for (int num : nums) {
			if (beforeNum != num) nums[idx++] = num;
			beforeNum = num;
		}
		
		// 3. λ°˜ν™˜
		return idx;
	}
}

회고

  • HashMap을 μ‚¬μš©ν•˜λ©΄ λŒ€λΆ€λΆ„μ˜ λ¬Έμ œκ°€ ν•΄κ²°λ˜λ‹€ λ³΄λ‹ˆ λ¬΄λΆ„λ³„ν•˜κ²Œ μ‚¬μš©μ„ ν•˜κ²Œ λ˜λŠ” 것 κ°™λ‹€.
  • 그런데 HashMap을 μ‚¬μš©ν•˜λ©΄ 값을 λ„£κ³  μ‘°νšŒν•˜λŠ” λΆ€λΆ„μ—μ„œ λΆˆν•„μš”ν•œ μ‹œκ°„ μ†Œμš”κ°€ λ°œμƒν•˜λ‹€ λ³΄λ‹ˆ κΌ­ ν•„μš”ν•œ κ²½μš°μ—λ§Œ μ‚¬μš©ν•  수 μžˆλ„λ‘ ν•΄μ•Όκ² λ‹€.Β 
  • 값을 μ‘°νšŒν•˜λŠ” λΆ€λΆ„μ—μ„œ μ‹œκ°„ μ†Œμš”κ°€ λ°œμƒν•˜λŠ” μ΄μœ λŠ” HashCode 연산이 λ°œμƒν•˜κΈ° λ•Œλ¬Έμ΄λ‹€.
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