All credits go to @Penghuan who thought of using the pairs wisely.

This method seems to be a bit simpler to understand, since it doesn't start with the left half and stays a little bit closer to the conventional solutions.

```
public static 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];
}
```