[Swift]LeetCode268. 缺失数字 | Missing Number

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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?

---

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

---

28ms

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

---

24ms

class Solution {
    func missingNumber(_ nums: [Int]) -> Int {
        var sum = 0 
        var max = 0
        var i = 0
        while i < nums.count {
            max = max + i
            sum = sum + nums[i]
            i = i + 1
        }
        return max - sum + i
    }
}

---

32ms：

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

class Solution {
    func missingNumber(_ nums: [Int]) -> Int {
        let count = nums.count
        var sum = count + (count * (count - 1)) / 2
        
        for i in 0..<count {
            sum -= nums[i]
        }
        return sum
    }
}