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【力扣100】4.移动零

article 2024/9/23 1:26:30

题目链接
我的题解：

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        # 思路是先计算共有几个0，然后remove几次，再末位加几个0
        count=0
        for i in nums:
            if i ==0:
                count=count+1
        for j in range(0,count):
            nums.remove(0)
            nums.append(0)

我觉得这个方法太蠢了
想一下双指针的方法

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        # fast指针遇见0向后挪
        # slow指针遇见0跟fast交换
        fast=0
        slow=0
        for fast in range(len(nums)):
            if nums[fast]==0:
                fast+=1
            else:
                nums[slow]=nums[fast]
                fast+=1
                slow+=1
        for i in range(slow,len(nums)):
            nums[i]=0

双指针,最后补0

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