Problem
Given an array A
of non-negative integers, half of the integers in A are odd, and half of the integers are even.
Sort the array so that whenever A[i]
is odd, i
is odd; and whenever A[i]
is even, i
is even.
You may return any answer array that satisfies this condition.
Example 1:
Input: [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.
Note:
2 <= A.length <= 20000
A.length % 2 == 0
0 <= A[i] <= 1000
Solution 1: Brute Force
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: vector<int> sortArrayByParityII(vector<int>& A) { vector<int> evens; vector<int> odds; for (int a : A) if (a % 2 == 0) evens.push_back(a); else odds.push_back(a); auto it1 = begin(evens); auto it2 = begin(odds); for (int i = 0; i < A.size(); ++i) { A[i] = *it1++; swap(it1, it2); } return A; } }; |
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