# 27. Remove Element (E)

Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` [**in-place**](https://en.wikipedia.org/wiki/In-place_algorithm). The relative order of the elements may be changed.

Since it is impossible to change the length of the array in some languages, you must instead have the result be placed in the **first part** of the array `nums`. More formally, if there are `k` elements after removing the duplicates, then the first `k` elements of `nums` should hold the final result. It does not matter what you leave beyond the first `k` elements.

Return `k` *after placing the final result in the first* `k` *slots of* `nums`.

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.

**Custom Judge:**

The judge will test your solution with the following code:

```
int[] nums = [...]; // Input array
int val = ...; // Value to remove
int[] expectedNums = [...]; // The expected answer with correct length.
                            // It is sorted with no values equaling val.

int k = removeElement(nums, val); // Calls your implementation

assert k == expectedNums.length;
sort(nums, 0, k); // Sort the first k elements of nums
for (int i = 0; i < actualLength; i++) {
    assert nums[i] == expectedNums[i];
}
```

If all assertions pass, then your solution will be **accepted**.

&#x20;

**Example 1:**

```
Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2,_,_]
Explanation: Your function should return k = 2, with the first two elements of nums being 2.
It does not matter what you leave beyond the returned k (hence they are underscores).
```

**Example 2:**

```
Input: nums = [0,1,2,2,3,0,4,2], val = 2
Output: 5, nums = [0,1,4,0,3,_,_,_]
Explanation: Your function should return k = 5, with the first five elements of nums containing 0, 0, 1, 3, and 4.
Note that the five elements can be returned in any order.
It does not matter what you leave beyond the returned k (hence they are underscores).
```

&#x20;

**Constraints:**

* `0 <= nums.length <= 100`
* `0 <= nums[i] <= 50`
* `0 <= val <= 100`

### Solution:

**Version 1:**

题目要求我们把 `nums` 中所有值为 `val` 的元素原地删除，依然需要使用 [双指针技巧](https://labuladong.github.io/algo/2/21/57/) 中的快慢指针：

如果 `fast` 遇到需要去除的元素，则直接跳过，否则就告诉 `slow` 指针，并让 `slow` 前进一步。

这和前面说到的数组去重问题解法思路是完全一样的，就不画 GIF 了，直接看代码：

```java
int removeElement(int[] nums, int val) {
    int fast = 0, slow = 0;
    while (fast < nums.length) {
        if (nums[fast] != val) {
            nums[slow] = nums[fast];
            slow++;
        }
        fast++;
    }
    return slow;
}
```

**注意这里和有序数组去重的解法有一个重要不同**，我们这里是先给 `nums[slow]` 赋值然后再给 `slow++`，这样可以保证 `nums[0..slow-1]` 是不包含值为 `val` 的元素的，最后的结果数组长度就是 `slow`。

**Version 2:**

1. Initialize `index` to 0, which represents the current position for the next non-target element.
2. Iterate through each element of the input array using the `i` pointer.
3. For each element `nums[i]`, check if it is equal to the target value.
   * If `nums[i]` is not equal to `val`, it means it is a non-target element.
   * Set `nums[index]` to `nums[i]` to store the non-target element at the current `index` position.
   * Increment `index` by 1 to move to the next position for the next non-target element.
4. Continue this process until all elements in the array have been processed.
5. Finally, return the value of `index`, which represents the length of the modified array.

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


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