540. Single element in ordered array

subject

540. A single element in an ordered array

The main idea of the topic

Give you an ordered array of integers only , Each of these elements will appear twice , Only one number will appear once .

Please find and return the number that only appears once .

The solution you design must meet O(log n) Time complexity and O(1) Spatial complexity .

Examples

Data scale

Ideas 1

Consider the simplest way first ： Because other elements appear twice , Only one element will appear once , Then the XOR value of the number that appears twice is 0, Then the XOR value of all numbers is the value of the element that appears only once .

Time complexity $O(n)$

Code 1

class Solution {
    
public:
    int singleNonDuplicate(vector<int>& nums) {
    
        int ans=0;
        for(int i=0;i<nums.size();i++){
    
            ans^=nums[i];
        }
        return ans;
    }
};

Ideas 2

Consider how to $O(logn)$ Realization ： The subscript corresponding to the first of the paired elements must be even , The subscript corresponding to the second of the paired elements must be odd . Then consider the existence of a single element , If the subscript of a single element is $x$, So subscript $x$ Before （ On the left ） The position of still satisfies the above conclusion , And subscripts $x$ after （ On the right ） Due to $x$ Insertion , Cause the conclusion to flip . So you can use dichotomy to judge contraction .

Time complexity $O(logn)$

Code 2

class Solution {
    
public:
    int singleNonDuplicate(vector<int>& nums) {
    
        int l=0,r=nums.size()-1,ans=0;
        while(l<r){
    
            int mid=(l+r)/2;
            if(nums[mid]==nums[mid^1]){
    
                l=mid+1;
            }
            else{
    
                r=mid;
                ans=mid;
            }
        }
        return nums[r];
    }
};