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set-mismatch.py
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set-mismatch.py
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# Time: O(n)
# Space: O(1)
# The set S originally contains numbers from 1 to n.
# But unfortunately, due to the data error, one of the numbers
# in the set got duplicated to another number in the set, which results
# in repetition of one number and loss of another number.
#
# Given an array nums representing the data status of this set after the error.
# Your task is to firstly find the number occurs twice and then find the number
# that is missing. Return them in the form of an array.
#
# Example 1:
# Input: nums = [1,2,2,4]
# Output: [2,3]
# Note:
# The given array size will in the range [2, 10000].
# The given array's numbers won't have any order.
class Solution(object):
def findErrorNums(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
x_xor_y = 0
for i in xrange(len(nums)):
x_xor_y ^= nums[i] ^ (i+1)
bit = x_xor_y & ~(x_xor_y-1)
result = [0] * 2
for i, num in enumerate(nums):
result[bool(num & bit)] ^= num
result[bool((i+1) & bit)] ^= i+1
if result[0] not in nums:
result[0], result[1] = result[1], result[0]
return result
# Time: O(n)
# Space: O(1)
class Solution2(object):
def findErrorNums(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
result = [0] * 2
for i in nums:
if nums[abs(i)-1] < 0:
result[0] = abs(i)
else:
nums[abs(i)-1] *= -1
for i in xrange(len(nums)):
if nums[i] > 0:
result[1] = i+1
else:
nums[i] *= -1
return result
# Time: O(n)
# Space: O(1)
class Solution3(object):
def findErrorNums(self, nums):
"""
:type nums: List[int]
:rtype: List[int]
"""
N = len(nums)
x_minus_y = sum(nums) - N*(N+1)//2
x_plus_y = (sum(x*x for x in nums) - N*(N+1)*(2*N+1)/6) // x_minus_y
return (x_plus_y+x_minus_y) // 2, (x_plus_y-x_minus_y) // 2