LeetCode 15 3Sum 解题思路和python代码

news2024/10/6 19:26:11

题目：
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.

Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

Constraints:

3 <= nums.length <= 3000
-105 <= nums[i] <= 105

解题思路：
首先对数组进行排序，有助于使用双指针方法快速找到合适的三元组。
对于每个元素nums[i]，将它作为第一个数，然后使用双指针方法在 i 之后的数组部分寻找另外两个数，使它们的和等于0。
左指针从 i+1 开始，右指针从数组末尾开始，逐步向中间收缩。根据三者之和的结果，调整左右指针的位置。
为了避免重复的三元组，遍历时需要跳过相同的元素。

class Solution(object):
    def threeSum(self, nums):
        """
        :type nums: List[int]
        :rtype: List[List[int]]
        """
        result = []
        nums.sort()

        print("nums: ", nums)

        for i in range(len(nums)):
            print("i: ", i)

            # 如果当前元素和前一个相同，跳过，避免重复三元组
            if i > 0 and nums[i] == nums[i-1]:
                continue
            
            left, right = i+1, len(nums)-1

            while left < right:
                print("nums[i]: ", nums[i])
                print("nums[left]: ", nums[left])
                print("nums[right]: ", nums[right])
                total = nums[i] + nums[left] + nums[right]
                print("total: ", total)

                if total == 0:
                    result.append([nums[i], nums[left], nums[right]])

                    while left < right and nums[left] == nums[left+1]:
                        left+=1
                    while left < right and nums[right] == nums[right-1]:
                        right-=1
                    
                    left+=1
                    right-=1
                
                elif total < 0:
                    left += 1
                else:
                    right -= 1
        
        return result