268. Missing Number

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Description

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.length
• 1 <= n <= 104
• 0 <= nums[i] <= n
• All the numbers of nums are unique.

Solutions

Python3

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        res = len(nums)
        for i, v in enumerate(nums):
            res ^= (i ^ v)
        return res

Java

• XOR

class Solution {
    public int missingNumber(int[] nums) {
        int res = nums.length;
        for (int i = 0, n = res; i < n; ++i) {
            res ^= (i ^ nums[i]);
        }
        return res;
    }
}

• Math

class Solution {
    public int missingNumber(int[] nums) {
        int res = nums.length;
        for (int i = 0, n = res; i < n; ++i) {
            res += (i - nums[i]);
        }
        return res;
    }
}

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