[Swift]LeetCode268. 缺失数字 | Missing Number

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 中没有出现在序列中的那个数。

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:
你的算法应具有线性时间复杂度。你能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少的那个数字。
 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 }

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 }

 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 }

求出从0~n的累加和，减去数组整体的和，那么由于数组内每个数字不相同，其差就是缺少的那个数字

 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 }

原文地址：https://www.cnblogs.com/strengthen/p/9756490.html