# Remove Element

## Description

Given an array *nums* and a value *val*, remove all instances of that value [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) and return the new length.

Do not allocate extra space for another array, you must do this by **modifying the input array** [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm) 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:

```java
// 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]);
}
```

## **Code**

```java
public int removeElement(int[] nums, int val) {
    // 记录后面元素要移动的距离
    int moveLen = 0;
    for (int i = 0; i < nums.length; i++) {
        if (moveLen != 0) {
            nums[i - moveLen] = nums[i];
        }
        if (nums[i] == val) {
            moveLen++;
        }
    }
    return nums.length - moveLen;
}
```