comments | difficulty | edit_url |
---|---|---|
true |
Medium |
Write a function to swap a number in place (that is, without temporary vari ables).
Example:
Input: numbers = [1,2] Output: [2,1]
Note:
numbers.length == 2
We can use the XOR operation
The XOR operation has the following three properties:
- Any number XORed with
$0$ remains unchanged, i.e.,$a \oplus 0=a$ . - Any number XORed with itself results in
$0$ , i.e.,$a \oplus a=0$ . - The XOR operation satisfies the commutative and associative laws, i.e.,
$a \oplus b \oplus a=b \oplus a \oplus a=b \oplus (a \oplus a)=b \oplus 0=b$ .
Therefore, we can perform the following operations on two numbers
-
$a=a \oplus b$ , now$a$ stores the XOR result of the two numbers; -
$b=a \oplus b$ , now$b$ stores the original value of$a$ ; -
$a=a \oplus b$ , now$a$ stores the original value of$b$ ;
In this way, we can swap two numbers without using a temporary variable.
The time complexity is
class Solution:
def swapNumbers(self, numbers: List[int]) -> List[int]:
numbers[0] ^= numbers[1]
numbers[1] ^= numbers[0]
numbers[0] ^= numbers[1]
return numbers
class Solution {
public int[] swapNumbers(int[] numbers) {
numbers[0] ^= numbers[1];
numbers[1] ^= numbers[0];
numbers[0] ^= numbers[1];
return numbers;
}
}
class Solution {
public:
vector<int> swapNumbers(vector<int>& numbers) {
numbers[0] ^= numbers[1];
numbers[1] ^= numbers[0];
numbers[0] ^= numbers[1];
return numbers;
}
};
func swapNumbers(numbers []int) []int {
numbers[0] ^= numbers[1]
numbers[1] ^= numbers[0]
numbers[0] ^= numbers[1]
return numbers
}
function swapNumbers(numbers: number[]): number[] {
numbers[0] ^= numbers[1];
numbers[1] ^= numbers[0];
numbers[0] ^= numbers[1];
return numbers;
}
class Solution {
func swapNumbers(_ numbers: [Int]) -> [Int] {
var numbers = numbers
numbers[0] ^= numbers[1]
numbers[1] ^= numbers[0]
numbers[0] ^= numbers[1]
return numbers
}
}