title: 540. Single Element in a Sorted Array
tags:
- Binary Search
categories: leetcode
comments: false
int ret = 0;
for(int a:nums) ret^=a;
option 1 - *Binary Search
class Solution {
public:
int singleNonDuplicate(vector<int>& nums) {
int n = nums.size();
// 一定是奇數長度
// 如果mid索引是偶數代表索引前面的數字應該都是兩兩一對,如果不是mid索引的值會與前一索引值相同,則落單的數必定在前面,反之相反。
int l = 0, r = n-1;
while(l<r){
int mid = l +(r-l)/2;
if (( mid % 2 == 1 && nums[mid] != nums[mid + 1]) || ( mid % 2 == 0 && nums[mid] == nums[mid + 1])) l = mid + 1;
else r= mid;
}
return nums[l];
}
};
- time complexity
O(logn)
- space complexity
O(1)