链接

https://leetcode-cn.com/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof/

前言

题目

输入一个整数数组，实现一个函数来调整该数组中数字的顺序，使得所有奇数位于数组的前半部分，所有偶数位于数组的后半部分。
示例

输入：nums = [1,2,3,4]
输出：[1,3,2,4] 
注：[3,1,2,4] 也是正确的答案之一。

限制：

0 <= nums.length <= 50000
1 <= nums[i] <= 10000

关键

本人思路

调用sort函数，通过条件x%2==0排序

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        nums.sort(key=lambda x:x%2==0)
        return nums

思路1

辅助数组

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        odd = []
        even = []
        for i in nums:
            if i%2==0:
                even.append(i)
            else:
                odd.append(i)
        return odd+even

思路2

首尾指针
首指针找偶数，尾指针找奇数，找到了就互换
注意跳出条件

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        left = 0
        right = len(nums)-1
        while left <= right:
            while left <= right and nums[left]%2 == 1:
                left += 1
            while left <= right and nums[right]%2 ==0:
                right -= 1
            if left > right:
                break
            nums[left], nums[right] = nums[right], nums[left]
        return nums

思路3

快慢指针
fast找奇数，找到了就和slow互换

class Solution:
    def exchange(self, nums: List[int]) -> List[int]:
        slow = fast = 0
        while fast < len(nums):
            if nums[fast]%2 == 1:
                nums[slow], nums[fast] = nums[fast], nums[slow]
                slow +=1
            fast +=1
        return nums