922. Sort Array By Parity II - Easy
문제
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.
제한사항
- 2 <= nums.length <= 2 * 104
- nums.length is even.
- Half of the integers in nums are even.
- 0 <= nums[i] <= 1000
입출력 예
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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.
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Example 2:
Input: nums = [2,3]
Output: [2,3]
풀이
- Two Pointer
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class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
int i = 0;
int j = 1;
while (i < nums.size() && j < nums.size()) {
if ((nums[i] & 1) == 0) {
i += 2;
} else if ((nums[j] & 1) == 1) {
j += 2;
} else {
std::swap(nums[i], nums[j]);
}
}
return nums;
}
};