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.nums are even.0 <= nums[i] <= 1000
Follow Up: Could you solve it in-place?
class Solution {
public:
vector<int> sortArrayByParityII(vector<int>& nums) {
int lt=0, rt=1, sz=nums.size();
while(1){
while(lt<sz && nums[lt]%2==lt%2)
lt+=2;
while(rt<sz && nums[rt]%2==rt%2)
rt+=2;
cout << lt<<"\t"<<rt<<"\n";
if(lt<sz && rt<sz){
int temp = nums[lt];
nums[lt] = nums[rt];
nums[rt] = temp;
}
else break;
}
return nums;
}
};