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This article mainly shows you "how LeetCode finds out numbers that appear only once". The content is simple and easy to understand, and the organization is clear. I hope it can help you solve your doubts. Let Xiaobian lead you to study and learn this article "how LeetCode finds out numbers that appear only once".

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Title Description

Given an array of non-empty integers where only one number appears once and the rest three times, find the number that appears only once. For example, input [3, 4, 5, 4, 3, 3], output 5.

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problem-solving

Idea 1: Set Difference

Slice the set to generate three sets and find the missing numbers for the sets that differ from the other two sets. For example, the set [2, 2, 3, 2] is sorted first to obtain [2, 2, 2, 3], and the sets [2, 3],[2], and [2] are generated by slicing, and the element differences between the sets are compared to obtain the output result 3.

class Solution: def singleNumber(self, nums: List[int]) -> int: nums.sort() a=len(set(nums[::2])) b=len(set(nums[1::2])) if a==b: return list(set(nums[::3]) - set(nums[2::3]))[0] else: return list(set(numbers [::3]) - set(numbers [1::3]))[0] Idea 2: Bit operation

By using the set method, the space-time complexity is O(N), and the bit operation can reduce the space complexity to O(1). In LeetCode brush problem DAY 5: only appear once in the number of "exclusive OR" in fact is the same state appears twice will return to zero, that is, 0->1->10=0 change, in this question is to let the same state appear three times to zero, that is, 00->01->10->11=00, it can be seen here need to use two bits for state recording.

It is important to note here that two and one are calculated separately. For one, when two=1, the next one is 0 regardless of the input, when two=0, input 1 is one=~one, input 0 is one unchanged. For two, depending on the state of one after change, when one new state is 1, two is 0, when one new state is 0, input 1 two=~two, input 0 is unchanged. Because three occurrences return to zero, one ends up with a value that only occurs once.

class Solution: def singleNumber(self, nums: List[int]) -> int: one,two = 0,0 for i in nums: one = one^i & ~two two = two^i & ~one return one Of course, traversing the set, recording the number of occurrences of each number through the hash table is also possible, and the space-time complexity is also O(N). The code is consistent with the idea of Day 5, so I won't repeat it here. That's all for "How LeetCode finds numbers that only appear once." Thanks for reading! I believe that everyone has a certain understanding, hope to share the content to help everyone, if you still want to learn more knowledge, welcome to pay attention to the industry information channel!

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