Question

link

Stats

Frequency	4
Diffficulty	1
Adjusted Difficulty	1
Time to use	----------

Ratings/Color = 1(white) 2(lime) 3(yellow) 4/5(red)

Solution

Similar to the question [LeetCode 26] Remove Duplicates From Sorted Array. Use 2 pointers.

Let’s play a game

The code I gave is 21 lines. It’s too long. Can we solve this problem with less code?

Sure we can! I have a 10-line version:

public class Solution {
    public int removeElement(int[] A, int elem) {
        int left = 0, right = 0;
        while (right < A.length) {
            if (A[right] == elem) right++;
            else A[left++] = A[right++];
        }
        return left;
    }
}

Now for a moment I thought the above code is the most concise, until I read this blog. The code is:

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

OK game over. Look at the standard answer below. Happy Leetcoding!

My code

public class Solution {
    public int removeElement(int[] A, int elem) {
        if (A == null || A.length == 0) {
            return 0;
        }
        int len = A.length;
        int left = 0;
        int right = 0;
        while (right < len) {
            // skip all instances of elem 
            while (right < len && A[right] == elem) {
                right++;
            }
            if (right == len) {
                break;
            }
            A[left++] = A[right++];
        }
        return left;
    }
}