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
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int l = 0, r = nums.size() - 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (nums[mid] >= target) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return nums[l] == target ? l : -1;
    }
};
