10/5 每日一題(找出數組中任意三個元素的sum=0, 且 i!=j!=k)

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

 class Solution:

    def threeSum(self, nums: List[int]) -> List[List[int]]:        
        nums.sort()#進行排序後使用雙指針找三元素
        res=set()
        n=len(nums)
        for i in range(n-2):
            #避免重複
            if i > 0 and nums[i] == nums[i-1]:
                continue
            left,right = i + 1, n - 1
            target = -nums[i]
                
            while left < right:
                s = nums[left]+nums[right] 
                
                if s == target:
                    data=(nums[i],nums[left],nums[right])
                    res.add(data)
                    #指針繼續移動
                    #進一步檢查下一項
                    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 s> target: #'表示太大'左移
                    right-=1
                else:
                    left += 1
        
        return list(res)