leetcode - 27. Remove Element-Toy模板网

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Description

Given an integer array nums and an integer val, remove all occurrences of val in nums in-place. The order of the elements may be changed. Then return the number of elements in nums which are not equal to val.

Consider the number of elements in nums which are not equal to val be k, to get accepted, you need to do the following things:

Change the array nums such that the first k elements of nums contain the elements which are not equal to val. The remaining elements of nums are not important as well as the size of nums.
Return k.

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

Description

Use two pointers, one from left and the other from right. Since we need to put all the numbers that are not equal to val, we iterate the list and compare the left and right every time. When the left equals to val, we swap the left and right, then move backwards the right, or otherwise we move forwards left. The left pointer actually points to all the numbers that are not equal to val, while the right pointer actually points to all numbers that are equal to val. So when left is larger than right, we have iterated all the numbers.

Time complexity: $o (n)$
Space complexity: $o (1)$ 文章来源地址https://www.toymoban.com/news/detail-490415.html

Code

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        left, right = 0, len(nums) - 1
        while left <= right:
            if nums[left] == val:
                nums[left], nums[right] = nums[right], nums[left]
                right -= 1
            else:
                left += 1
        return left