[Swift]LeetCode268. 缺失数字 | Missing Number

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（shanqingyongzhi）
➤博客园地址：山青咏芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：http://www.javashuo.com/article/p-fmdoiqcb-kz.html 
➤若是连接不是山青咏芝的博客园地址，则多是爬取做者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持做者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

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

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?

---

给定一个包含 0, 1, 2, ..., n 中 n 个数的序列，找出 0 .. n 中没有出如今序列中的那个数。

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:
你的算法应具备线性时间复杂度。你可否仅使用额外常数空间来实现?

---

 1 class Solution {
 2     func missingNumber(_ nums: [Int]) -> Int {
 3         //利用异或运算，将数组全体内容与0~n进行异或,
 4         //根据异或运算的性质可知最后结果为缺乏的那个数字。
 5         var result:Int = nums.count
 6         for i in 0..<nums.count
 7         {
 8             result ^= i ^ nums[i]
 9         }
10         return result        
11     }
12 }

---

28ms

1 class Solution {
2     func missingNumber(_ nums: [Int]) -> Int {
3         var h = (nums.count+1)*nums.count/2
4         for i in 0..<nums.count {
5            h -= nums[i]
6         }
7         return h
8     }
9 }

---

24ms

 1 class Solution {
 2     func missingNumber(_ nums: [Int]) -> Int {
 3         var sum = 0 
 4         var max = 0
 5         var i = 0
 6         while i < nums.count {
 7             max = max + i
 8             sum = sum + nums[i]
 9             i = i + 1
10         }
11         return max - sum + i
12     }
13 }

---

32ms：

求出从0~n的累加和，减去数组总体的和，那么因为数组内每一个数字不相同，其差就是缺乏的那个数字

 1 class Solution {
 2     func missingNumber(_ nums: [Int]) -> Int {
 3         let count = nums.count
 4         var sum = count + (count * (count - 1)) / 2
 5         
 6         for i in 0..<count {
 7             sum -= nums[i]
 8         }
 9         return sum
10     }
11 }