Binary Search
<p>Given an array of integers <code>nums</code> which is sorted in ascending order, and an integer <code>target</code>, write a function to search <code>target</code> in <code>nums</code>. If <code>target</code> exists, then return its index. Otherwise, return <code>-1</code>.</p>

<p>You must write an algorithm with <code>O(log n)</code> runtime complexity.</p>

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

<pre>
<strong>Input:</strong> nums = [-1,0,3,5,9,12], target = 9
<strong>Output:</strong> 4
<strong>Explanation:</strong> 9 exists in nums and its index is 4
</pre>

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

<pre>
<strong>Input:</strong> nums = [-1,0,3,5,9,12], target = 2
<strong>Output:</strong> -1
<strong>Explanation:</strong> 2 does not exist in nums so return -1
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>4</sup> &lt; nums[i], target &lt; 10<sup>4</sup></code></li>
	<li>All the integers in <code>nums</code> are <strong>unique</strong>.</li>
	<li><code>nums</code> is sorted in ascending order.</li>
</ul>
Array; Binary Search
func search(nums []int, target int) int {
	l, r := 0, len(nums)-1
	for l < r {
		mid := (l + r) >> 1
		if nums[mid] >= target {
			r = mid
		} else {
			l = mid + 1
		}
	}
	if nums[l] == target {
		return l
	}
	return -1
}
