27. Remove Element

Posted on | Edited on | Views:

LeetCode

Approach 1: Two Pointers, Time complexity: O(2*N)=O(N).

1
2
3
4
5
6
7
8
9
10
11
12
class Solution {
    public int removeElement(int[] nums, int val) {
        if(nums.length==0) return 0;
        int j=-1;
        for(int i=0;i<nums.length;i++){
            while(i<nums.length&&nums[i]==val) i++;
            if(i>=nums.length) break;
            nums[++j]=nums[i];
        }
        return ++j;
    }
}

Approach 2: Two Pointers - when elements to remove are rare, Time complexity: O(N)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
class Solution {
    public int removeElement(int[] nums, int val) {
        int i=0,n=nums.length;
        while(i<n){
            if(nums[i]==val){
                nums[i]=nums[n-1];
                n--;
            }else{
                i++;
            }
        }
        return n;
    }
}

0%