Given a sorted array consisting of only integers where every element appears twice except for one element which appears once. Find this single element that appears only once.

Example 1:
Input: [1,1,2,3,3,4,4,8,8]
Output: 2

Example 2:
Input: [3,3,7,7,10,11,11]
Output: 10

Solution：Binary Search

思路：
如果当前数字出现两次的话，我们可以通过数组的长度跟当前位置的关系，计算出右边和当前数字不同的数字的总个数，如果是偶数个，说明落单数左半边，反之则在右半边。
Time Complexity: O(logN) Space Complexity: O(1)

Solution Code：

class Solution {
    public int singleNonDuplicate(int[] nums) {
        int start = 0, end = nums.length - 1;

        while (start < end) {
            // We want the first element of the middle pair,
            // which should be at an even index if the left part is sorted.
            // Example:
            // Index: 0 1 2 3 4 5 6
            // Array: 1 1 3 3 4 8 8
            //            ^
            int mid = (start + end) / 2;
            if (mid % 2 == 1) mid--;

            // We didn't find a pair. The single element must be on the left.
            // (pipes mean start & end)
            // Example: |0 1 1 3 3 6 6|
            //               ^ ^
            // Next:    |0 1 1|3 3 6 6
            if (nums[mid] != nums[mid + 1]) end = mid;

            // We found a pair. The single element must be on the right.
            // Example: |1 1 3 3 5 6 6|
            //               ^ ^
            // Next:     1 1 3 3|5 6 6|
            else start = mid + 2;
        }

        // 'start' should always be at the beginning of a pair.
        // When 'start > end', start must be the single element.
        return nums[start];
    }
}