Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
Example 1:
Input: nums = [3,0,1] Output: 2 Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:
Input: nums = [0,1] Output: 2 Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:
Input: nums = [9,6,4,2,3,5,7,0,1] Output: 8 Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
Example 4:
Input: nums = [0] Output: 1 Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.
Constraints:
n == nums.length1 <= n <= 1040 <= nums[i] <= nnums are unique.struct Solution;
impl Solution {
fn missing_number(nums: Vec<i32>) -> i32 {
let mut xor: i32 = 0;
let n = nums.len();
for n in 0..=n {
xor ^= n as i32;
}
for n in nums {
xor ^= n;
}
xor
}
}
#[test]
fn test() {
let nums = vec![9, 6, 4, 2, 3, 5, 7, 0, 1];
assert_eq!(Solution::missing_number(nums), 8);
}