This article is participating in Python Theme Month. See the link for details

describe

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]; }Copy the code

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

Example 1:

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).Copy the code

Example 2:

Input: nums =,1,2,2,3,0,4,2 [0], 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).Copy the code

Note:

0 <= nums.length <= 100
0 <= nums[i] <= 50
0 <= val <= 100
Copy the code

parsing

Val = val; val = val; val = val; val = val; val = val; val = val Question 26 is similar to the following:

  • Initialize a counter count to record how many Val’s to drop
  • If the current element nums[I] is equal to val, the counter count is incremented by one. If the current element nums[I] is not equal, nums[I] is assigned to nums[i-count], which is equivalent to advancing all legal elements by count
  • At the end of the loop, len(nums)-count is returned as the number of elements after removing all val.

answer

class Solution(object):
    def removeElement(self, nums, val):
        """
        :type nums: List[int]
        :type val: int
        :rtype: int
        """
        if len(nums) == 0: return  0
        count = 0
        for i in range(len(nums)):
            if nums[i] == val:
                count += 1
            else:
                nums[i-count] = nums[i]
        return len(nums)-count

        	      
		
Copy the code

The results

Given the given list in the Python online submissions. Memory Usage: 10 ms Given in the Python online submissions for Remove elements.Copy the code

parsing

Here’s another idea:

  • Initialize count to represent the index of a non-val element
  • If nums[I] is not equal to val, nums[I] is assigned to nums[count] and count is incremented by one. The non-val elements in nums are then banished to positions in front of nums, one by one, as count increments
  • At the end of the loop, count is the previous number of non-val elements.

answer

class Solution(object): def removeElement(self, nums, val): """ :type nums: List[int] :type val: int :rtype: int """ if len(nums) == 0: return 0 count = 0 for i in range(len(nums)): if nums[i]! =val: nums[count] = nums[i] count+=1 return countCopy the code

The results

Given in the Python online submission. Memory Usage: 10000 ms Submissions in Python online submissions for Remove elements.Copy the code

Original link: leetcode.com/problems/re…

Your support is my biggest motivation