题目:原题链接(简单)
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
---|---|---|---|
Ans 1 (Python) | O(1) | 460ms (5.71%) | |
Ans 2 (Python) | O(n) | O(n) | 40ms (85.03%) |
LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。
解法一(两层迭代):
def moveZeroes(self, nums: List[int]) -> None:
size = len(nums)
i = 0
zero = 0
while i + zero < size:
n = nums[i]
if n == 0:
for j in range(i, len(nums) - 1):
nums[j] = nums[j + 1]
nums[-1] = 0
zero += 1
else:
i += 1
解法二(数组存储非0位):
def moveZeroes(self, nums: List[int]) -> None:
sites = []
for i in range(len(nums)):
n = nums[i]
if n != 0:
sites.append(n)
index = 0
for n in sites:
nums[index] = n
index += 1
for i in range(len(sites), len(nums)):
nums[i] = 0