LeetCode之268.Missing number

1. Description

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. Follow Up

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

3. Test Cases

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.

4. Solution

4.1

如果没有想到什么奇淫巧计，那就老老实实写个答案先。

class Solution {
    
    public int missingNumber(int[] nums) {
    
        Set<Integer> numSet = new HashSet<Integer>();
        
        for(int num : nums) {
    
            numSet.add(num);
        }
        
        int expectedNumCount = nums.length + 1;
        
        for (int number=0; number<expectedNumCount; number++) {
    
            if (!numSet.contains(number)) {
    
                return number;
            }
        }
        
        return -1;
    }
}

这样做的效率显然不高。

4.3

这里就可以使用特殊的技能了。
异或操作是计算机中一个神奇的操作，可以简单的理解为相同为0，不同为1.如果使用0和任何数异或，那么结果都是任何数。
这里，由于我们知道数组中只缺少了一个数字，而且数字值的大小是数组长达+1的大小为最大值。那么我们可以看到数组[3,0,1],数组长度为3，对应的数字为0,1，2,3四个数字，但是少了一个。考虑到对应的下标，分别为0,1，2，这样我们可以使用数组的下标与数组字异或，再多比上数组长度+1，即
3 ^ 0 ^ 3 ^ 1 ^ 0 ^ 2 ^ 1 = 2,即缺失的就是2。

class Solution {
    
    public int missingNumber(int[] nums) {
    
        int missing = nums.length;   
        for (int i = 0; i < nums.length; i++) {
    
            missing ^= i ^nums[i];
        }
        return missing;
    }
}