摘要：

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note: The solution set must not contain duplicate triplets.

For example, given array S = [-1, 0, 1, 2, -1, -4],

A solution set is:

[
  [-1, 0, 1],
  [-1, -1, 2]
]

問題分析：
 對數組實現3個值的求和，可以分解為兩個數值的和；問題需要考慮到過濾重復的解決方案和考慮到算法復雜度；
 

 1 class Solution(object):
 2     def threeSum(self, nums):
 3         """
 4         :type nums: List[int]
 5         :rtype: List[List[int]]
 6         """
 7         nums.sort() # 根據python的內置列表排序，實現排序
 8         result = list()            
 9         if len(nums) < 3:
10             return result #
 
 判斷，如果輸入列表小於3，直接返回空值
11         for i in range(len(nums) - 2): # 只能遍歷到len(nums)-2,否則越界
12             if i > 0 and nums[i] == nums[i-1]:
13                 continue
14             left = i + 1
15             right = len(nums) - 1
16             while left < right:
17                 cur_sum = nums[left] + nums[right]
 
18                 target_sum = 0 - nums[i]
19                 if cur_sum == target_sum:
20                     temp = [nums[i],nums[left],nums[right]]
21                     result.append(temp)
22                     left +=1 
23                     right -= 1
24                     while left < right and nums[left]==nums[left-1]:
25                         left +=1 
26                     while left < right and nums[right] == nums[right+1]:
27                         right -= 1
28                 elif cur_sum < target_sum:
29                     left += 1
30                 else:
31                     right -= 1
32            return result

代碼解析：

　　第一想法：對數組按照i,j,k三個指針指向進行遍歷，時間復雜度為O(n³),當數組長度很大時，運算速度過慢，造成無法運算

　　第二想法：通過搜索查看別人答案，上面的代碼實現時間復雜度為O(nlogn);邏輯為，遍歷一次數組，不重復遍歷。轉化為求2sum的問題

leetcode 3sum question