Sort Array by Parity II Problem & Solution
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.
See the sort array by parity ii problem on LeetCode.
C++ Solution
#pragma GCC optimize("Ofast")
#pragma GCC optimization("unroll-loops")
static const int _=[](){std::ios::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);return 0;}();
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
int even = 0;
int odd = 1;
int i = 0;
while (i < nums.size()) {
if (nums[i] % 2 == 0 && i % 2 == 1) {
swap(nums[i], nums[even]);
even += 2;
} else if (nums[i] % 2 == 1 && i % 2 == 0) {
swap(nums[i], nums[odd]);
odd += 2;
} else {
++i;
}
}
return nums;
}
};