15. 3Sum
Đề Bài
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
Thuật Toán & Kỹ Thuật
⏱️ Thời gian
O(n log n)
💾 Không gian
O(n)
Lời Giải
Python
0015-3sum.py
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
n = len(nums)
nums.sort()
ans = set()
for i in range(n - 2):
l, r = i + 1, n - 1
target = -nums[i]
while l < r:
total = nums[l] + nums[r]
if total == target:
ans.add((nums[i], nums[l], nums[r]))
r -= 1
l += 1
elif total > target:
r -= 1
else:
l += 1
return ans