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This article mainly introduces LeetCode how to solve the problem of the sum of three numbers, the article introduces in great detail, has a certain reference value, interested friends must read it!

one

Topic description

Given an integer array nums, it is judged whether there are three elements in nums, such that a + b + c = 0. If there is no return [], if there is, return all the answers that meet the criteria and do not repeat. For example, input [- 1mem0pence2jormae1mae4] returns [[- 1jie0recovery1], [- 1maimel 1mem2]], and returns [] if you enter [- 3jue 3].

two

Solve a problem

This question needs two predictions: 1, when the length of the array is less than 3, the direct output []; 2, the array is sorted first, if the current number is the same as the previous one, the results will be the same, just skip it.

Idea 1: hash table

This problem is to find the three elements that meet the conditions. When the first element an is fixed, the problem is transformed into the problem of finding b and c to make the sum of-a, which is consistent with the problem in LeetCode exercise DAY 8: the sum of two numbers, so it can also be solved by hash table.

Class Solution: def threeSum (self, nums: List [int])-> list [int]: if len (nums) 0 and nums [I] = = nums [I-1]: continue h_map = {} target =-nums [I] for j in range (iTun1) Len (nums): if target-nums [j] in h_map: a.append (sorted ([nums [I], nums [j], target- nums [j])) h _ map [nums [j]] = j return list (set ([tuple (t) for t in a]))

Idea 2: double pointers

When the array is sorted and the first element an is fixed, the problem is the same as the problem in LeetCode exercise DAY 9: the sum of two numbers II, and can be solved by double pointer method.

Class Solution: def threeSum (self, nums: List [int])-> list [int]: if len (nums) 0 and nums [I] = nums [I-1]: continue x = len (nums)-1 target =-nums [I] while x

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