Decode XORed Permutation
<p>There is an integer array <code>perm</code> that is a permutation of the first <code>n</code> positive integers, where <code>n</code> is always <strong>odd</strong>.</p>

<p>It was encoded into another integer array <code>encoded</code> of length <code>n - 1</code>, such that <code>encoded[i] = perm[i] XOR perm[i + 1]</code>. For example, if <code>perm = [1,3,2]</code>, then <code>encoded = [2,1]</code>.</p>

<p>Given the <code>encoded</code> array, return <em>the original array</em> <code>perm</code>. It is guaranteed that the answer exists and is unique.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> encoded = [3,1]
<strong>Output:</strong> [1,2,3]
<strong>Explanation:</strong> If perm = [1,2,3], then encoded = [1 XOR 2,2 XOR 3] = [3,1]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> encoded = [6,5,4,6]
<strong>Output:</strong> [2,4,1,5,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>3 &lt;= n &lt;&nbsp;10<sup>5</sup></code></li>
	<li><code>n</code>&nbsp;is odd.</li>
	<li><code>encoded.length == n - 1</code></li>
</ul>
Bit Manipulation; Array
class Solution {
    public int[] decode(int[] encoded) {
        int n = encoded.length + 1;
        int a = 0, b = 0;
        for (int i = 0; i < n - 1; i += 2) {
            a ^= encoded[i];
        }
        for (int i = 1; i <= n; ++i) {
            b ^= i;
        }
        int[] perm = new int[n];
        perm[n - 1] = a ^ b;
        for (int i = n - 2; i >= 0; --i) {
            perm[i] = encoded[i] ^ perm[i + 1];
        }
        return perm;
    }
}