LeetCode解题报告（307)-- 268. Missing Number

Problem

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:

1
2
3

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:

1
2
3

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:

1
2
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:

1
2
3

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.

Analysis

这道题目和上一题有点类似，但是有一点不同，这里是长度为n的数组，但是数字是[0,n]一共n+1个，所以没办法与下标一一对应。但是这里只需要求出缺失的值，所以很简单的思路就是把前n项求和，减去数组中的元素，剩下的结果就是缺失的数字了。

Solution

无。

Code

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int sum = 0;
        int n = nums.size();
        for (auto &num: nums) {
            sum += num;
        }
        return 0.5 * n * (n + 1) - sum;
    }
};

Summary

这道题目实际上和数组并没有很大关系，巧用数学公式来求解即可。这道题目的分享到这里，感谢你的支持！