LeetCode 268. Missing Number - 求和、异或

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?

题意：0～n中间少了一个数

思路1：每个位置放了一个数，但是有一个位置没有放。采用求和的方法，可以把没有放的这个数找出来。

class Solution(object):
def missingNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
sum = len(nums)
for i in range(len(nums)):
sum += i
sum -= nums[i]
return sum

思路2：异或运算 a^b^b=a，这个运算满足交换律，大致思想同上。

class Solution(object):
def missingNumber(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
xor = len(nums)
for i in range(len(nums)):
xor ^= i ^ nums[i]
return xor