[LeetCode]27. Remove Element

Given an array nums and a value val, remove all instances of that
value in-place and return the new length.

Do not allocate extra space for another array, you must do this by
modifying the input array in-place with 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 containing 0, 1, 3, 0, and 4.

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

It doesn't matter what values are set beyond the returned length.
Clarification:

Confused why the returned value is an integer but your answer is an
array?

Note that the input array is passed in by reference, which means
modification to the input array will be known to the caller as well.

Internally you can think of this:

// nums is passed in by reference. (i.e., without making a copy) int
len = removeElement(nums, val);

// any modification to nums in your function would be known by the
caller. // using the length returned by your function, it prints the
first len elements. for (int i = 0; i < len; i++) {

print(nums[i]); }

注意不僅返回長度，也要改變數組

public int removeElement(int[] nums, int val) {
        int len=0;
        for(int i:nums){
            if(i==val) continue;
            len++;
            nums[len-1]=i;
        }
        return len;
}