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Formatted question description: https://leetcode.ca/all/905.html
905. Sort Array By Parity (Easy)
Given an array A
of non-negative integers, return an array consisting of all the even elements of A
, followed by all the odd elements of A
.
You may return any answer array that satisfies this condition.
Example 1:
Input: [3,1,2,4] Output: [2,4,3,1] The outputs [4,2,3,1], [2,4,1,3], and [4,2,1,3] would also be accepted.
Note:
1 <= A.length <= 5000
0 <= A[i] <= 5000
Solution 1. STL
// OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
sort(begin(A), end(A), [](int a, int b) { return a % 2 < b % 2; });
return A;
}
};
Solution 2. Two Pointers
// OJ: https://leetcode.com/problems/sort-array-by-parity/
// Time: O(N)
// Space: O(1)
class Solution {
public:
vector<int> sortArrayByParity(vector<int>& A) {
for (int i = 0, j = A.size() - 1; i < j; ++i, --j) {
while (i < j && A[i] % 2 == 0) ++i;
while (i < j && A[j] % 2 != 0) --j;
if (i < j) swap(A[i], A[j]);
}
return A;
}
};
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class Solution { public int[] sortArrayByParity(int[] A) { if (A == null || A.length < 2) return A; int length = A.length; for (int i = 1; i < length; i++) { int num = A[i]; if (num % 2 != 0) continue; int insertIndex = 0; for (int j = i - 1; j >= 0; j--) { if (A[j] % 2 == 0) { insertIndex = j + 1; break; } else A[j + 1] = A[j]; } A[insertIndex] = num; } return A; } } ############ class Solution { public int[] sortArrayByParity(int[] nums) { for (int i = 0, j = nums.length - 1; i < j;) { if (nums[i] % 2 == 1) { int t = nums[i]; nums[i] = nums[j]; nums[j] = t; --j; } else { ++i; } } return nums; } }
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// OJ: https://leetcode.com/problems/sort-array-by-parity/ // Time: O(NlogN) // Space: O(1) class Solution { public: vector<int> sortArrayByParity(vector<int>& A) { sort(begin(A), end(A), [](int a, int b) { return a % 2 < b % 2; }); return A; } };
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class Solution: def sortArrayByParity(self, nums: List[int]) -> List[int]: i, j = 0, len(nums) - 1 while i < j: if nums[i] & 1: nums[i], nums[j] = nums[j], nums[i] j -= 1 else: i += 1 return nums ############ class Solution(object): def sortArrayByParity(self, A): """ :type A: List[int] :rtype: List[int] """ return sorted(A, key = lambda x : x % 2)
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func sortArrayByParity(nums []int) []int { for i, j := 0, len(nums)-1; i < j; { if nums[i]%2 == 1 { nums[i], nums[j] = nums[j], nums[i] j-- } else { i++ } } return nums }
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/** * @param {number[]} nums * @return {number[]} */ var sortArrayByParity = function (nums) { for (let i = 0, j = nums.length - 1; i < j; ) { if (nums[i] & 1) { [nums[i], nums[j]] = [nums[j], nums[i]]; --j; } else { ++i; } } return nums; };
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impl Solution { pub fn sort_array_by_parity(mut nums: Vec<i32>) -> Vec<i32> { let (mut l, mut r) = (0, nums.len() - 1); while l < r { while l < r && nums[l] & 1 == 0 { l += 1; } while l < r && nums[r] & 1 == 1 { r -= 1; } nums.swap(l, r); } nums } }