[LeetCode] #27. Remove Element (repeat : 0)

 

지정된 배열의 요소를 삭제하는 문제이다. 배열의 인덱스만 변경해 간단하게 풀었다. 

 

 

Description

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).

 

#Solution 

class Solution {
    public int removeElement(int[] nums, int val) {
        int newSize = 0;    
        for(int i=0;i<nums.length;i++){
            if(nums[i] != val ){
                nums[newSize] = nums[i];
                newSize++;
            }
        }
        return newSize;
    }
}

In terms of space complexity, it is O(1) because it uses only limited constant variables like newSize and nums.

In terms of time complexity, it is O(n) because the main operation involves in iterating the nums array. 'n' represents number of elements in nums array.