题目描述

题目链接：268. Missing Number

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.

例子

例子 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.

例子 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.

例子 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.

Follow Up

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

Constraints

• n == nums.length
• 1 <= n <= 10^4
• 0 <= nums[i] <= n
• All the numbers of nums are unique.

解题思路

这道题和 [Leetcode]136. Single Number(C++) 类似，同样可以利用异或的性质求解，在 136 中已经给定了成对的数字需要找出只有一个数字，这里我们可以先将 1 ～ n 的数字异或成一个数字，再和数字中的 n 个数字异或这样相同的数字会抵消掉，剩下的就是数组中缺失的数了。代码如下：

#include <vector>

class Solution {
public:
    int missingNumber(std::vector<int>& nums) {
        int mask = 0;
        for (int i = 0; i <= nums.size(); ++i) {
            mask ^= i;
        }

        for (int num : nums) {
            mask ^= num;
        }

        return mask;
    }
};

• 时间复杂度: O(n)
• 空间复杂度: O(1)

GitHub 代码同步地址： 268.MissingNumber.cpp

其他题目： GitHub: Leetcode-C++-Solution 博客： Leetcode-Solutions