Given an arraynumsand a valueval, remove all instances of that valuein-placeand return the new length.

Do not allocate extra space for another array, you must do this bymodifying the input arrayin-placewith O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new length.

Example 1:

Given nums = [3,2,2,3], val = 3,

Your function should return length = 2, with the first two elements of nums being 2.

It doesn't matter what you leave beyond the returned length.

Example 2:

Given nums = [0,1,2,2,3,0,4,2], val = 2,

Your function should return length = 5, with the first five elements of nums
 
 containing0, 1, 3, 0, and4.

Note that the order of those five elements can be arbitrary.

It doesn't matter what values are set beyondthe returned length.

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

總結：

Since we are asked to return the length of the new array and don't need to worry about vals beyond new length.

We must delete vals in place. So we used two pointers all starts from 0, i means new array's index, j means old array's index.

If current val != val, we do nums[i++] = nums[j], otherwise we skip modification of index i to make all elements != val.