https://leetcode.com/problems/wiggle-subsequence/
- 注意这里的判断条件
class Solution:
def wiggleMaxLength(self, nums: List[int]) -> int:
if len(nums) < 2:
return len(nums)
pre = 0
res = 1
for i in range(1, len(nums)):
diff = nums[i] - nums[i-1]
if pre * diff <= 0 and diff != 0:
res += 1
pre = diff
return res时间复杂度：O()
空间复杂度：O()
class Solution:
def wiggleSort(self, nums: List[int]) -> None:
# 1. If i is even, then nums[i] <= nums[i - 1].
# 2. If i is odd, then nums[i] >= nums[i - 1].
for i in range(1, len(nums)):
if i & 1 == 0 and nums[i] > nums[i - 1] or i & 1 == 1 and nums[i] < nums[i - 1]:
nums[i], nums[i - 1] = nums[i - 1], nums[i]时间复杂度：O(n)
空间复杂度：O(1)
class Solution:
def wiggleSort(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
arr = sorted(nums)
for i in range(1, len(nums), 2):
nums[i] = arr.pop()
for i in range(0, len(nums), 2):
nums[i] = arr.pop()- 结合
n-largest与wiggle sort I的方法