[LeetCode][Python]268. Missing N

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

For example,
Given nums = [0, 1, 3] return 2.

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

先介绍一下Python中的异或操作：

异或运算：A与B不同为1时，A、B的预算结果才为1，否则为0 （运算符：^）

异或交换原理: 数字A异或B两次，就得到A。而B被A异或两次，就得到B

任意数与自己异或操作都是0，任意数与0异或操作都得到本身。

思路:是再创建一个n个元素的列表，与提供的相加，问题就转化为在一个列表中，除了一个元素只出现一次之外，其余的都出现两次，然后找出这个只出现一次的元素。

class Solution(object):
    def missingNumber(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        temp = nums + list(range(len(nums)+1))
        result = 0
        for ele in temp:
            result = result^ele
        return result


if __name__ == '__main__':
    sol = Solution()
    nums = [0, 1, 3]
    print sol.missingNumber(nums)
    nums = [0]
    print sol.missingNumber(nums)
    nums = [1]
    print sol.missingNumber(nums)

    nums = [0, 1, 2, 3, 4, 5, 6]
    print sol.missingNumber(nums)

别人家的思路：

1. 数字A异或B两次，就得到A

    def missingNumber2(self, nums):
        result = 0
        for i in range(len(nums)+1):
            result = result^i
        for num in nums:
            result ^= num
        return result

2. Sort and then Binary Search

    def missingNumber3(self, nums):
        nums.sort()
        left = 0
        right = len(nums)
        while(left <=right):
            mid = (left+right)/2
            if nums[mid] > mid:
                right = mid - 1
            else:
                left = mid + 1
        return left