[LeetCode] 27. Remove Element - Java

 

Remove Element - LeetCode

 

문제

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val 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 elements which are not equal to val. 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:

int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                            // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
    assert nums[i] == expectedNums[i];
}

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

 

Example 1:

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

Example 2:

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

 

Constraints:

• 0 <= nums.length <= 100
• 0 <= nums[i] <= 50
• 0 <= val <= 100

 

주어진 배열 nums 에서 val 과 일치하는 원소들을 찾아 모두 제거하는 문제이다.

nums 배열 자체에서 원소를 제거하고, 제거된 뒤 nums에 남은 원소의 개수 k 를 반환한다.

 

---

 

내 풀이

class Solution {
    public int removeElement(int[] nums, int val) {

        int k = 0;
        for(int i=0; i<nums.length; i++){
           if(nums[i] != val){
               nums[k] = nums[i];
               k++;
           }
        }
        return k;
    }
}

제거한 뒤 남아있는 원소들을 제외하고 그 뒤에 값들은 중요하지 않기 때문에

단순하게 val 과 일치하지 않을 때 nums[k] 에 해당 값을 넣어주면서 k를 증가시키면 된다.