文章目录
一、题目
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?
二、题解
使用哈希表的方法解决
cpp
class Solution {
public:
int missingNumber(vector<int>& nums) {
unordered_map<int,int> map;
int n = nums.size();
for(int i = 0;i < n;i++) map[nums[i]] = 1;
for(int i = 0;i <= n;i++){
if(!map.count(i)) return i;
}
return 0;
}
};