Deleting Items From an Array

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LeetCode 27
LeetCode 26

LeetCode 27: Remove Element

Given an array nums and a value val, remove all instances of that value in-place and return the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory.

The order of elements can be changed. It doesn't matter what you leave beyond the new 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 a modification to the input array will be known to the caller as well.

Internally you can think of this:

// 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]);
}

Example 1:

Input: nums = [3,2,2,3], val = 3
Output: 2, nums = [2,2]
Explanation: 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. For example if you return 2 with nums = [2,2,3,3] or nums = [2,2,0,0], your answer will be accepted.

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 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.

Constraints:

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

我的解法: 利用 BubbleSort

q27_BubbleSort.pyview raw
class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        l = len(nums)
        count = 0
        for i in range(l-1):
            swapped = False
            for j in range(l-1-i):
                if nums[j] == val:
                    if nums[j] != nums[j+1]:
                        nums[j], nums[j+1] = nums[j+1], nums[j]
                        swapped = True
            if not swapped:
                break
        for v in nums:
            if v != val:
                count += 1
        return count

我的解法: 利用 QuickSort

q27_QuickSort.pyview raw
class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        if len(nums) == 0:
            return 0
        else:
            i = 0
            j = len(nums)-1
            while i != j:
                while nums[i] != val and i < j:
                    i += 1
                while nums[j] == val and i < j:
                    j -= 1
                if i < j:
                    nums[i], nums[j] = nums[j], nums[i]
        if nums[i] == val:
            return i
        else:
            return i+1

其他解法 1

q27解.pyview raw
class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        count = 0
        for i in range(len(nums)):
            if nums[i] != val:
                nums[count] = nums[i]
                count +=1
        return count

其他解法 2

q27解2.pyview raw
class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
		i = 0
		n = len(nums)
		while i < n:
			if nums[i] == val:
				nums[i] = nums[n-1]
				n -= 1
	    	else:
				i += 1
		return n

LeetCode 26: Remove Duplicates from Sorted Array

Given a sorted array nums, remove the duplicates in-place such that each element appears only once and returns the new length.

Do not allocate extra space for another array, you must do this by modifying the input array in-place with O(1) extra memory. Clarification: