leetcode-268-easy

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.

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?

思路一：观察数组的分布范围，刚好是 [0, n)，把数组全部相加，再减去[0, n]，由于数组少了一个值，所以结果一定是所求解的负数，因为少了那个值

public int missingNumber(int[] nums) {
    int x = -nums.length;
    for (int i = 0; i < nums.length; i++) {
        x += nums[i];
        x -= i;
    }

    return -x;
}

思路二：用 set 统计，然后遍历也是一种思路，但是需要额外的空间