leetcode 540. Single Element in a Sorted Array

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
Note: Your solution should run in O(log n) time and O(1) space.

题目大意:一个有序的数列，有一个数出现一次，其他的数出现两次。求这个出现一次的数

思路：二分，首先序列是有序的，所以相邻的两个数要么相等，要么不相等。如果不相等，那会是哪个区间有single num呢 我们再判断区间(r-mid+1)是不是奇数
1.区间mid~r是奇数，并且nums[mid+1]和nums[mid]不等,说明single num出现在l~mid。
2.区间mid~r是偶数，并且nums[mid+1]和nums[mid]相等，说明single num出现在l~mid-1
3.区间mid~r是奇数，并且nums[mid+1]和nums[mid]相等，说明single num出现在mid~r
4.区间mid~r是偶数，并且nums[mid+1]和nums[mid]不等，single num出现在mid~r

代码如下：

class Solution {
public:

    int singleNonDuplicate(vector<int>& nums) {
        int l = 0;
        int r = nums.size()-1;
        int mid = 0;
        while (l < r) {
            mid = (l+r)/2;
            if (nums[mid+1] != nums[mid]) mid = mid+1;
            if ((r - mid + 1)%2) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return nums[r];
    }
};