540. Single Element in a Sorted Array

Time O(logN) Space O(1)
100.00 %
class Solution {
    public int singleNonDuplicate(int[] nums) {
        int start = 0;
        int end = (nums.length - 1) / 2;
        // find the first lead index that nums[index] != nums[index + 1]
        while (start < end) {
            int mid = start + (end - start) / 2;
            if (nums[mid * 2] == nums[mid * 2 + 1]) {
                start = mid + 1;
            } else {
                end = mid;
            }
        }
        return nums[start * 2];
    }
}