LeetCode.Single Number II

Problem:

Given a non-empty array of integers, every element appears three times except for one, which appears exactly once. Find that single one.

Note:

Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

 

Example 1:

Input: [2,2,3,2]

Output: 3

 

Example 2:

Input: [0,1,0,1,0,1,99]

Output: 99

 

-Summary-

Solved by using dictionary. 

1. for each number in the nums list, we save the key(as the number) in the dictionary and count. 

2. return the key with a count value equals 1.

#from collections import defaultdict
class Solution:
    def singleNumber(self, nums: List[int]) -> int:
        res = 0
        for i in range(32):
            count = 0
            for n in nums:
                count += (n >> i) & 1
            rem = count % 3
            if i == 31 and rem:  # must handle the negative case
                res -= 1 << 31
            else:
                res |= rem << i
        
        return res
        '''
        #Use extra memory of dict()
        numsDict = defaultdict(int)
        
        for num in nums:
            numsDict[num] += 1
        
        for key in numsDict:
            if numsDict[key] == 1:
                return key
        '''

Learned bit manipulation way to solve without using extra memory from the below discussion.

leetcode.com/problems/single-number-ii/discuss/43367/Python-Easy-Understand-Solution-using-32-bit-counters

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