LeetCode #268 Missing Number 缺失数字

Description:
Given an array containing n distinct numbers taken from 0, 1, 2, ..., n, find the one that is missing from the array.

Example :
Example 1:
Input: [3,0,1]
Output: 2

Example 2:
Input: [9,6,4,2,3,5,7,0,1]
Output: 8

Note:
Your algorithm should run in linear runtime complexity. Could you implement it using only constant extra space complexity?

题目描述:
给定一个包含 0, 1, 2, ..., n 中 n 个数的序列，找出 0 .. n 中没有出现在序列中的那个数。

示例 :
示例 1:
输入: [3,0,1]
输出: 2

示例 2:
输入: [9,6,4,2,3,5,7,0,1]
输出: 8

说明：
你的算法应具有线性时间复杂度。你能否仅使用额外常数空间来实现?

思路:

1. 参考LeetCode #136 Single Number 只出现一次的数字的思路, 利用异或
2. 求和, 与通项公式求取的值相差的数就是丢失的数
   时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int result = nums.size();
        for (int i = 0; i < nums.size(); i++) {
            result ^= nums[i];
            result ^= i;
        }
        return result;
    }
};

Java:

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

Python:

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        return (len(nums) ** 2 + len(nums)) // 2 - sum(nums)