# 21.调整数组顺序使奇数位于偶数前面

## 题目描述

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。

提示：

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

来源：力扣（LeetCode） 链接：<https://leetcode-cn.com/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof> 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

## 解法一

最容易想到的

```swift
func exchange(_ nums: [Int]) -> [Int] {
    var temp: [Int] = []
    nums.forEach {
        if $0 & 1 == 1 {
            temp.insert($0, at: 0)
        }
        else {
            temp.append($0)
        }
    }
    return temp
}
```

## 解法二：差速双指针

配上这个动图食用更佳[来源](https://leetcode-cn.com/problems/diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof/solution/ti-jie-shou-wei-shuang-zhi-zhen-kuai-man-shuang-zh/)

![差速双指针](/files/-Me4Ovduz7n1fKufEZ-C)

```swift
func exchange1(_ nums: [Int]) -> [Int] {
    var fastP: Int = 0
    var slowP: Int = 0
    var temp = nums

    while fastP < nums.count {
        if temp[fastP] & 1 == 1 {
            temp.swapAt(fastP, slowP)
            slowP += 1
        }
        fastP += 1
    }

    return temp
}
```


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