LeetCode 268 Missing Number

LeetCode

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.

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.

Constraints:

n == nums.length 1 <= n <= 10<sup>4</sup>
0 <= nums[i] <= n All the numbers of nums are unique.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

题目大意：

数组缺失一个数，所有数应该在[0, n]内，求缺失数

排序法解题思路：

N/A

解题步骤：

N/A

注意事项：

Python代码：

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def canPermutePalindrome(self, s: str) -> bool:
	char_to_count = collections.Counter(s)
	return False if len([count for count in char_to_count.values() if count % 2 == 1]) > 1 else True

算法分析：

时间复杂度为O(nlogn)，空间复杂度O(1)

异或法解题思路II：

高斯原理

Python代码：

def missingNumber2(self, nums: List[int]) -> int:
	res = len(nums) # remember
	for i, n in enumerate(nums):
		res ^= i ^ n
	return res

算法分析：

时间复杂度为O(n)，空间复杂度O(1)

数学法解题思路III：

高斯原理

Python代码：

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def missingNumber3(self, nums: List[int]) -> int:
	n = len(nums)
	return (0 + n) * (n + 1) // 2 - sum(nums)

算法分析：

时间复杂度为O(n)，空间复杂度O(1)