268. Missing Number

1. Question

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.

2. Examples

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.

Example 4:

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.

3. 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?

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/missing-number 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

5.1. 排序遍历

class Solution {
  public int missingNumber(int[] nums) {
    Arrays.sort(nums);
    int ans = 0;

    for(int i = 0; i < nums.length; i++) {
      if (nums[i] != i) {
        return i;
      }
    }

    return nums.length;
  }
}

5.2. Set排序遍历

class Solution {
  public int missingNumber(int[] nums) {
    HashSet<Integer> set = new HashSet<>();

    int ans = nums.length;

    for(int i : nums) {
      set.add(i);
    }

    for(int i = 0; i < nums.length; i++) {
      if (!set.contains(i)) {
        ans = i;
        break;
      }
    }

    return ans;
  }
}

5.3. 数学法

等差数列

class Solution {
  public int missingNumber(int[] nums) {
    int len = nums.length;

    // 等差数列
    int sum = len * (len + 1) / 2;

    // 求出数组中所有数字的和
    int res = Arrays.stream(nums).sum();

    return sum - res;

  }
}