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268.missing-number

Statement

Metadata
  • Link: 丢失的数字
  • Difficulty: Easy
  • Tag: 位运算 数组 哈希表 数学 排序

给定一个包含 [0, n] 中 n 个数的数组 nums ，找出 [0, n] 这个范围内没有出现在数组中的那个数。

 

示例 1：

输入：nums = [3,0,1]
输出：2
解释：n = 3，因为有 3 个数字，所以所有的数字都在范围 [0,3] 内。2 是丢失的数字，因为它没有出现在 nums 中。

示例 2：

输入：nums = [0,1]
输出：2
解释：n = 2，因为有 2 个数字，所以所有的数字都在范围 [0,2] 内。2 是丢失的数字，因为它没有出现在 nums 中。

示例 3：

输入：nums = [9,6,4,2,3,5,7,0,1]
输出：8
解释：n = 9，因为有 9 个数字，所以所有的数字都在范围 [0,9] 内。8 是丢失的数字，因为它没有出现在 nums 中。

示例 4：

输入：nums = [0]
输出：1
解释：n = 1，因为有 1 个数字，所以所有的数字都在范围 [0,1] 内。1 是丢失的数字，因为它没有出现在 nums 中。

 

提示：

  • n == nums.length
  • 1 <= n <= 104
  • 0 <= nums[i] <= n
  • nums 中的所有数字都 独一无二

 

进阶：你能否实现线性时间复杂度、仅使用额外常数空间的算法解决此问题?

Metadata
  • Link: Missing Number
  • Difficulty: Easy
  • Tag: Bit Manipulation Array Hash Table Math Sorting

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 <= 104
  • 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?

Solution

from typing import List


class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        nums = list(map(lambda x: x + 1, nums))
        nums.append(1)
        nums.append(1)
        n = len(nums)
        for i in range(n - 2):
            x = abs(nums[i])
            nums[x] *= -1

        for i in range(1, n):
            if nums[i] > 0:
                return i - 1
最后更新: October 11, 2023
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