Search in Rotated Sorted Array
<p>There is an integer array <code>nums</code> sorted in ascending order (with <strong>distinct</strong> values).</p>

<p>Prior to being passed to your function, <code>nums</code> is <strong>possibly rotated</strong> at an unknown pivot index <code>k</code> (<code>1 &lt;= k &lt; nums.length</code>) such that the resulting array is <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code> (<strong>0-indexed</strong>). For example, <code>[0,1,2,4,5,6,7]</code> might be rotated at pivot index <code>3</code> and become <code>[4,5,6,7,0,1,2]</code>.</p>

<p>Given the array <code>nums</code> <strong>after</strong> the possible rotation and an integer <code>target</code>, return <em>the index of </em><code>target</code><em> if it is in </em><code>nums</code><em>, or </em><code>-1</code><em> if it is not in </em><code>nums</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 = [4,5,6,7,0,1,2], target = 0
<strong>Output:</strong> 4
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [4,5,6,7,0,1,2], target = 3
<strong>Output:</strong> -1
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1], target = 0
<strong>Output:</strong> -1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li>All values of <code>nums</code> are <strong>unique</strong>.</li>
	<li><code>nums</code> is an ascending array that is possibly rotated.</li>
	<li><code>-10<sup>4</sup> &lt;= target &lt;= 10<sup>4</sup></code></li>
</ul>
Array; Binary Search
class Solution {
public:
    int search(vector<int>& nums, int target) {
        int n = nums.size();
        int left = 0, right = n - 1;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums[0] <= nums[mid]) {
                if (nums[0] <= target && target <= nums[mid])
                    right = mid;
                else
                    left = mid + 1;
            } else {
                if (nums[mid] < target && target <= nums[n - 1])
                    left = mid + 1;
                else
                    right = mid;
            }
        }
        return nums[left] == target ? left : -1;
    }
};