题目要求

给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。

请注意 ，必须在不复制数组的情况下原地对数组进行操作。

示例 1:
输入: nums = [0,1,0,3,12]
输出: [1,3,12,0,0]
示例 2:
输入: nums = [0]
输出: [0]

解析

双指针：
i：快指针，遍历整个数组 ；j:慢指针，指向待处理的零元素位置
注意： 执行快慢指针需满足i>j

图片示例

Implement code

• 方法1

class Solution:
    def moveZeroes(self, nums: list[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        # i: Fast pointer, travase the entire array
        # index: Slow pointer, an index value that points to zero 
        index = i = 0
        while i < len(nums):
            if nums[i] != 0:
                nums[index], nums[i]= nums[i], nums[index]
                index += 1
            i += 1

• 放法2

class Solution:
    def moveZeroes(self, nums: list[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        # i: Fast pointer, travase the entire array
        # j: Slow pointer, an index value that points to zero 
        j = 0
        for i in range(len(nums)):
            if nums[i] != 0:
                if i > j: # Need to execute speed point
                    nums[j] = nums[i]
                    nums[i] = 0
                j += 1   
        return nums  # return None, so  no need to write

算法评估

时间复杂度：O(n)
空间复杂度：O(1)