Given a non-empty array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
Example 1:
Input: [2,2,1] Output: 1
Example 2:
Input: [4,1,2,1,2] Output: 4
这一题很有意思,做应该是会做,但是用unordered_set 做的话,虽然是线性时间,但也是很慢。看了题解之后,才明白了数学的美妙之处。还要多积累知识,多发散思维。
Approach : Bit Manipulation
Concept
- If we take XOR of zero and some bit, it will return that bit
- a \oplus 0 = aa⊕0=a
- If we take XOR of two same bits, it will return 0
- a \oplus a = 0a⊕a=0
- a \oplus b \oplus a = (a \oplus a) \oplus b = 0 \oplus b = ba⊕b⊕a=(a⊕a)⊕b=0⊕b=b
So we can XOR all bits together to find the unique number.
Complexity Analysis
Time complexity : O(n)O(n). We only iterate through \text{nums}nums, so the time complexity is the number of elements in \text{nums}nums.
Space complexity : O(1)O(1).
class Solution {
public:
int singleNumber(vector<int>& nums) {
int a = 0;
for(auto num: nums)
a ^= num;
return a;
}
};