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15. 3 Sum ​

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
Related Topics
  • 数组
  • 双指针
  • 排序

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    • 思路一 ​

    本题比较 tricky 如何去除重复的元素。

    • 采用 x in array 判断重复直接超时
    • 采用 set 值为 tuple 后可以通过测试，但是效率还是依然低下
    python
    class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = set()
        nums.sort()

        size = len(nums)
        for i in range(size - 2):
            left = i + 1
            right = size - 1

            target = 0 - nums[i]

            while left < right:
                sum = nums[left] + nums[right]
                if sum == target:
                    temp = (nums[i], nums[left], nums[right])
                    if temp not in res:
                        res.add(temp)
                    left += 1
                    right -= 1
                elif sum > target:
                    right -= 1
                else:
                    left += 1

        return [[r[0], r[1], r[2]] for r in res]

    复杂度分析 ​

    • 时间复杂度O(N2)
    • 空间复杂度O(N)

    解法二 ​

    思路 ​

    判读重复的元组的开销，不如从源头直接跳过重复的元素

    py
    from typing import List
# leetcode submit region begin(Prohibit modification and deletion)
class Solution:
    def threeSum(self, nums: List[int]) -> List[List[int]]:
        res = []
        nums.sort()

        size = len(nums)
        for i in range(size - 2):

            # 跳过宠物
            if i > 0 and nums[i] == nums[i - 1]:
                continue

            left = i + 1
            right = size - 1

            target = 0 - nums[i]

            while left < right:
                sum = nums[left] + nums[right]
                if sum == target:
                    res.append([nums[i], nums[left], nums[right]])
                    left += 1
                    right -= 1
                    while left < right and nums[left] == nums[left - 1]:
                        left += 1
                    while left < right and nums[right] == nums[right + 1]:
                        right -= 1
                elif sum > target:
                    right -= 1
                else:
                    left += 1

        return res

        
# leetcode submit region end(Prohibit modification and deletion)

Solution().threeSum([-4,-2,-2,-2,0,1,2,2,2,3,3,4,4,6,6])

    复杂度分析 ​

    • 时间复杂度O(n2)
    • 空间复杂度O(N), N = 有效的元组数量