Class Solution
java.lang.Object
g0901_1000.s0922_sort_array_by_parity_ii.Solution
922 - Sort Array By Parity II.<p>Easy</p>
<p>Given an array of integers <code>nums</code>, half of the integers in <code>nums</code> are <strong>odd</strong> , and the other half are <strong>even</strong>.</p>
<p>Sort the array so that whenever <code>nums[i]</code> is odd, <code>i</code> is <strong>odd</strong> , and whenever <code>nums[i]</code> is even, <code>i</code> is <strong>even</strong>.</p>
<p>Return <em>any answer array that satisfies this condition</em>.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> nums = [4,2,5,7]</p>
<p><strong>Output:</strong> [4,5,2,7]</p>
<p><strong>Explanation:</strong> [4,7,2,5], [2,5,4,7], [2,7,4,5] would also have been accepted.</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> nums = [2,3]</p>
<p><strong>Output:</strong> [2,3]</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>2 <= nums.length <= 2 * 10<sup>4</sup></code></li>
<li><code>nums.length</code> is even.</li>
<li>Half of the integers in <code>nums</code> are even.</li>
<li><code>0 <= nums[i] <= 1000</code></li>
</ul>
<p><strong>Follow Up:</strong> Could you solve it in-place?</p>
-
Constructor Summary
Constructors -
Method Summary
-
Constructor Details
-
Solution
public Solution()
-
-
Method Details
-
sortArrayByParityII
public int[] sortArrayByParityII(int[] nums)
-