LeetCode-M-Single Element in a Sorted Array

題意

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.

解法

二分查找：

• 當前值與左值、右值均不相等，返回當前值
• 當前值與左值相等
  
  • 左邊元素個數爲偶數：low = mid + 1
  • 左邊元素個數爲奇數：high = mid
• 當前值與右值相等
  
  • 左邊元素個數爲偶數：low ＝ mid
  • 左邊元素個數爲奇數：high ＝ mid - 1

實現

class Solution {
public:
    int singleNonDuplicate(vector<int>& nums) {
        if(nums.size() <= 0) return -1;
        int low = 0;
        int high = nums.size() - 1;
        while(low + 2 < high){
            int mid = low + (high - low)/2;
            if(nums[mid-1] != nums[mid] && nums[mid+1] != nums[mid]) return nums[mid];
            else if(nums[mid-1] == nums[mid]){
                if((mid - low + 1) % 2 == 0) low = mid + 1;
                else high = mid;
            }else if(nums[mid+1] == nums[mid]){
                if((mid - low) % 2 == 0) low = mid;
                else high = mid - 1;
            }
        }
        if(low == high) return nums[low];
        else if(low + 1 == high) return -1;//非法值
        else{
            int mid = low + (high - low)/2;
            if(nums[low] == nums[mid]) return nums[high];
            else if(nums[high] == nums[mid]) return nums[low];
            else return -1;
        }

    }
};