Description
Given an array of integers nums, half of the integers in nums are odd, and the other half are even.
Sort the array so that whenever nums[i] is odd, i is odd, and whenever nums[i] is even, i is even.
Return any answer array that satisfies this condition.
Example 1:
Input: nums = [4,2,5,7]
Output: [4,5,2,7]
Explanation: [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.
Example 2:
Input: nums = [2,3]
Output: [2,3]
Constraints:
2 <= nums.length <= 2 * 104nums.length is even.- Half of the integers in
nums are even.
0 <= nums[i] <= 1000
Follow Up: Could you solve it in-place?
Solutions
Solution 1: Two Pointers
We use two pointers i and j to point to even and odd indices, respectively. Initially, i = 0 and j = 1.
When i points to an even index, if nums[i] is odd, we need to find an odd index j such that nums[j] is even, and then swap nums[i] and nums[j]. Continue traversing until i reaches the end of the array.
The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustJavaScript
class Solution:
def sortArrayByParityII(self, nums: List[int]) -> List[int]:
n, j = len(nums), 1
for i in range(0, n, 2):
if nums[i] % 2:
while nums[j] % 2:
j += 2
nums[i], nums[j] = nums[j], nums[i]
return nums(code-box)
class Solution {
public int[] sortArrayByParityII(int[] nums) {
for (int i = 0, j = 1; i < nums.length; i += 2) {
if (nums[i] % 2 == 1) {
while (nums[j] % 2 == 1) {
j += 2;
}
int t = nums[i];
nums[i] = nums[j];
nums[j] = t;
}
}
return nums;
}
}(code-box)
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
for (int i = 0, j = 1; i < nums.size(); i += 2) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
swap(nums[i], nums[j]);
}
}
return nums;
}
};(code-box)
func sortArrayByParityII(nums []int) []int {
for i, j := 0, 1; i < len(nums); i += 2 {
if nums[i]%2 == 1 {
for nums[j]%2 == 1 {
j += 2
}
nums[i], nums[j] = nums[j], nums[i]
}
}
return nums
}(code-box)
function sortArrayByParityII(nums: number[]): number[] {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
}
}
return nums;
}(code-box)
impl Solution {
pub fn sort_array_by_parity_ii(mut nums: Vec<i32>) -> Vec<i32> {
let n = nums.len();
let mut j = 1;
for i in (0..n).step_by(2) {
if nums[i] % 2 != 0 {
while nums[j] % 2 != 0 {
j += 2;
}
nums.swap(i, j);
}
}
nums
}
}(code-box)
/**
* @param {number[]} nums
* @return {number[]}
*/
var sortArrayByParityII = function (nums) {
for (let i = 0, j = 1; i < nums.length; i += 2) {
if (nums[i] % 2) {
while (nums[j] % 2) {
j += 2;
}
[nums[i], nums[j]] = [nums[j], nums[i]];
}
}
return nums;
};(code-box)
