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 = 0while i < len(nums):if nums[i] != 0:nums[index], nums[i]= nums[i], nums[index]index += 1i += 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 = 0for i in range(len(nums)):if nums[i] != 0:if i > j: # Need to execute speed pointnums[j] = nums[i]nums[i] = 0j += 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)