# 27.remove-element

## Statement

• Difficulty: Easy
• Tag: `数组` `双指针`

``````// nums 是以“引用”方式传递的。也就是说，不对实参作任何拷贝
int len = removeElement(nums, val);
// 在函数里修改输入数组对于调用者是可见的。
// 根据你的函数返回的长度, 它会打印出数组中 该长度范围内 的所有元素。
for (int i = 0; i < len; i++) {
print(nums[i]);
}
``````

``````输入：nums = [3,2,2,3], val = 3

``````

``````输入：nums = [0,1,2,2,3,0,4,2], val = 2

``````

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

• Difficulty: Easy
• Tag: `Array` `Two Pointers`

Given an integer array `nums` and an integer `val`, remove all occurrences of `val` in `nums` in-place. 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 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.

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).
``````

Constraints:

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

## Solution

``````from typing import List

class Solution:
def removeElement(self, nums: List[int], val: int) -> int:
n = len(nums)
j = 0
for i in range(n):
if nums[i] != val:
nums[j] = nums[i]
j += 1

return j
``````