Leetcode 283. move zero

Power button | 283. Move zero

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Method 1 ： Two traversal

Use two cycles , The first cycle records the non-zero part of the array and overwrites the original array . Record the number of zero elements in the array .

Second cycle from n-count Start , Assign zero to the element at the end of the array .

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        count = 0
        j = 0
        for i in range(0, n):
            if nums[i]==0:
                count += 1
            else:
                nums[j] = nums[i]
                j += 1
        for i in range(n-count, n):
            nums[i] = 0

Complete test code

from typing import List

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        n = len(nums)
        count = 0
        j = 0
        for i in range(0, n):
            if nums[i]==0:
                count += 1
            else:
                nums[j] = nums[i]
                j += 1
        for i in range(n-count, n):
            nums[i] = 0

class main:
    a = Solution()
    nums = [6,0,53,0,11,25,7]
    a.moveZeroes(nums)
    print(nums)


if __name__ == '__main__':
    main()

  Method 2 ： Double pointer

Set two pointers , At the beginning, they all point to the position 0 The element of . When encountering non-zero elements, the left and right pointers exchange elements , Then move both pointers to the right at the same time . When it encounters zero element, it only moves the right pointer , At this time, the left pointer stays at the position of the zero element . Until the right pointer finds a non-zero element, exchange elements again , Then move right at the same time .

When the left and right pointers point to the same position , Exchanging elements is equivalent to no operation .

When the pointer is found to move to the zero element , Always move the right pointer before switching , It means constantly looking for non-zero elements in the back to fill the front position .

    def moveZeroes(self, nums: List[int]) -> None:
        n = len(nums)
        left, right = 0, 0
        for right in range(0, n):
            if nums[right] == 0 :
                right += 1
            else:
                nums[right], nums[left] = nums[left], nums[right]
                right += 1
                left += 1

Or the right +=  1 Refer to the outside of the judgment statement

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        n = len(nums)
        left, right = 0, 0
        for right in range(0, n):
            if nums[right] != 0 :
                nums[right], nums[left] = nums[left], nums[right]
                left += 1
            right += 1