6/6 每日一題（移動0）

 Given an integer array nums, move all 0's to the end of it while maintaining the relative order of the non-zero elements.

Note that you must do this in-place without making a copy of the array.

 

Example 1:

Input: nums = [0,1,0,3,12]
Output: [1,3,12,0,0]

Example 2:

Input: nums = [0]
Output: [0]

 

Constraints:

• 1 <= nums.length <= 104
• -231 <= nums[i] <= 231 - 1

 

Follow up: Could you minimize the total number of operations done?

class Solution:

    def moveZeroes(self, nums: List[int]) -> None:

        """

        Do not return anything, modify nums in-place instead.

        """

        zero_num=nums.count(0)#計算有幾個0

        n=len(nums)#長度

        i=0

        while i <n-zero_num:

            if nums[i]==0:

                zero=nums.pop(i)

                nums.append(zero)

                #為了確保遺漏下一項是0的狀況,所以要再檢查一次

                i -= 1

            i +=1

-----參考解答

class Solution: def moveZeroes(self, nums: List[int]) -> None: zero_count = 0 # 計算遇到的 0 的個數 n = len(nums) for i in range(n): if nums[i] != 0: # 將非零元素向前移動 zero_count 步 nums[i - zero_count] = nums[i] else: zero_count += 1 # 將末尾的 zero_count 個元素設為 0 for i in range(n - zero_count, n): nums[i] = 0