題目敘述
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]
Constraints:
1 <= nums.length <= 50000 <= nums[i] <= 5000
題目翻譯
這題是有一個整數陣列 nums，要將其中偶數元素往前挪移，奇數往後排序。其中偶數元素間或奇數元素間的排序不要求。
解法解析
這題跟上一篇 Move Zeroes 有 78 成相似，所以可以互相參考一下。但因為不考慮元素間的排序，所以處理方式相對簡單。
解法範例
Go
Sort
func sortArrayByParity(nums []int) []int {
sort.Slice(nums, func(i, j int) bool {
return nums[i]%2 < nums[j]%2
})
return nums
}
Two Pointers
func sortArrayByParity(nums []int) []int {
even := []int{}
odd := []int{}
for _, v := range nums {
if v%2 == 0 {
even = append(even, v)
} else {
odd = append(odd, v)
}
}
return append(even, odd...)
}
In Place
func sortArrayByParity(nums []int) []int {
for i, j := 0, len(nums)-1; i < j; {
if nums[i]%2 > nums[j]%2 {
nums[i], nums[j] = nums[j], nums[i]
}
if nums[i]%2 == 0 {
i++
}
if nums[j]%2 == 1 {
j--
}
}
return nums
}
JavaScript
Sort
var sortArrayByParity = function (nums) {
nums.sort((a, b) => (a % 2) - (b % 2));
return nums;
};
Two Pointers
var sortArrayByParity = function (nums) {
const result = new Array();
nums.forEach((e) => {
if (e % 2 == 1) {
result.push(e);
} else {
result.unshift(e);
}
});
return result;
};
In Place
var sortArrayByParity = function (nums) {
let i = 0,
j = nums.length - 1;
while (i < j) {
if (nums[i] % 2 > nums[j] % 2) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}
if (nums[i] % 2 == 0) {
i++;
}
if (nums[j] % 2 == 1) {
j--;
}
}
return nums;
};
Kotlin
Sort
class Solution {
fun sortArrayByParity(nums: IntArray): IntArray {
return nums.sortedBy { it % 2 }.toIntArray()
}
}
Two Pointers
class Solution {
fun sortArrayByParity(nums: IntArray): IntArray {
return nums.filter { it % 2 == 0 }.toIntArray().plus(nums.filter { it % 2 == 1 })
}
}
In Place
class Solution {
fun sortArrayByParity(nums: IntArray): IntArray {
var i = 0
var j = nums.size - 1
while (i < j) {
if (nums[i] % 2 > nums[j] % 2) {
nums[i] = nums[j].also { nums[j] = nums[i] }
}
if (nums[i] % 2 == 0) {
i++
}
if (nums[j] % 2 == 1) {
j--
}
}
return nums
}
}
PHP
Sort
class Solution
{
function sortArrayByParity($nums)
{
usort($nums, function ($a, $b) {
return ($a % 2) - ($b % 2);
});
return $nums;
}
}
Two Pointers
class Solution
{
function sortArrayByParity($nums)
{
$even = [];
foreach ($nums as $num) {
if ($num % 2 === 0) {
$even[] = $num;
}
}
return array_merge($even, array_diff($nums, $even));
}
}
In Place
class Solution
{
function sortArrayByParity($nums)
{
$i = 0;
$j = count($nums) - 1;
while ($i < $j) {
if ($nums[$i] % 2 > $nums[$j] % 2) {
$tmp = $nums[$i];
$nums[$i] = $nums[$j];
$nums[$j] = $tmp;
}
if ($nums[$i] % 2 === 0) {
$i++;
}
if ($nums[$j] % 2 === 1) {
$j--;
}
}
return $nums;
}
}
Python
Sort
class Solution:
def sortArrayByParity(self, nums: List[int]) -> List[int]:
nums.sort(key=lambda x: x % 2)
return nums
Two Pointers
class Solution:
def sortArrayByParity(self, nums: List[int]) -> List[int]:
return [x for x in nums if x % 2 == 0] + [x for x in nums if x % 2 == 1]
In Place
class Solution:
def sortArrayByParity(self, nums: List[int]) -> List[int]:
i, j = 0, len(nums) - 1
while i < j:
if nums[i] % 2 > nums[j] % 2:
nums[i], nums[j] = nums[j], nums[i]
if nums[i] % 2 == 0:
i += 1
if nums[j] % 2 == 1:
j -= 1
return nums
Rust
Swift
Sort
class Solution {
func sortArrayByParity(_ nums: [Int]) -> [Int] {
return nums.sorted {
$0 % 2 < $1 % 2
}
}
}
Two Pointers
class Solution {
func sortArrayByParity(_ nums: [Int]) -> [Int] {
return nums.filter { $0 % 2 == 0 } + nums.filter { $0 % 2 == 1 }
}
}
In Place
class Solution {
func sortArrayByParity(_ nums: [Int]) -> [Int] {
var i: Int = 0
var j: Int = nums.count - 1
var nums: [Int] = nums
while i < j {
if nums[i] % 2 == 0 {
i += 1
} else {
nums.swapAt(i, j)
j -= 1
}
}
return nums
}
}