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
func decode(encoded []int) []int {
	n := len(encoded) + 1
	a, b := 0, 0
	for i := 0; i < n-1; i += 2 {
		a ^= encoded[i]
	}
	for i := 1; i <= n; i++ {
		b ^= i
	}
	perm := make([]int, n)
	perm[n-1] = a ^ b
	for i := n - 2; i >= 0; i-- {
		perm[i] = encoded[i] ^ perm[i+1]
	}
	return perm
}