本文已参与掘金创作者训练营第三期「高产更文」赛道，详情查看：掘力计划｜创作者训练营第三期正在进行，「写」出个人影响力
描述
Given an integer array nums, move all the even integers at the beginning of the array followed by all the odd integers.
Return any array that satisfies this condition.
Example 1:
Input: nums = [3,1,2,4]
Output: [2,4,3,1]
Explanation: The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Example 2:
Input: nums = [0]
Output: [0]
Note:
1 <= nums.length <= 5000
0 <= nums[i] <= 5000
解析
根据题意，就是给出了一个整数列表 nums ，要求将其中的全部偶数都放到全部的奇数前面。思路比较简单，直接在 nums 原列表上进行操作：
- 用 index 记录处理元素的索引，用 count 记录处理了多少个数字
- 如果 nums[index] 是奇数，就将其放到 nums 的末尾，同时 count 加一
- 如果 nums[index] 是偶数，则不发生变化，将 count 和 index 同时加一
- 遍历结束，返回 nums 即可
解答
class Solution(object):
def sortArrayByParity(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
L = len(nums)
index = 0
count = 0
while count<L:
if nums[index]%2==1:
nums.append(nums.pop(index))
else:
index += 1
count += 1
return nums
运行结果
Runtime: 72 ms, faster than 20.73% of Python online submissions for Sort Array By Parity.
Memory Usage: 14.3 MB, less than 74.48% of Python online submissions for Sort Array By Parity.
解析
另外可以直接在原列表中进行操作，这样不用占用额外的空间。因为偶数要放在前面，奇数放在后面，所以：
- 初始化最左边索引 l 和最右边索引 r
- 当 l<r 的时候，一直执行 while ，将前面的奇数和后面的偶数进行交换，并不断更新 l 和 r 的值
- 遍历结束得到的新的 nums 即为答案
解答
class Solution(object):
def sortArrayByParity(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
l = 0
r = len(nums)-1
while l<r:
if nums[l]%2==1 and nums[r]%2==0:
nums[l], nums[r] = nums[r], nums[l]
if nums[l]%2==0: l+=1
if nums[r]%2==1: r-=1
return nums
运行结果
Runtime: 68 ms, faster than 41.56% of Python online submissions for Sort Array By Parity.
Memory Usage: 14.5 MB, less than 16.04% of Python online submissions for Sort Array By Parity.
解析
还可以使用 python 的内置函数 sort ，自定义比较器即可。
解答
class Solution(object):
def sortArrayByParity(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
nums.sort(key = lambda x:x%2)
return nums
运行结果
Runtime: 65 ms, faster than 41.98% of Python online submissions for Sort Array By Parity.
Memory Usage: 14.2 MB, less than 93.12% of Python online submissions for Sort Array By Parity.
原题链接：leetcode.com/problems/so…
您的支持是我最大的动力
