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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:
<pre>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. </pre>
Example 2:
<pre>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. </pre>
Example 3:
<pre>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. </pre>
Constraints:
n == nums.length1 <= n <= 10<sup>4</sup>0 <= nums[i] <= n- All the numbers of
numsare unique.
Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?
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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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3def 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：
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5def 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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3def 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)
