题目要求是给定数组nums，求无重复三元组，使三元组相加为0。
Given array nums = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
以下算法复杂度O(n^2)且直接除去重复三元组
class Solution:
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
nums = sorted(nums) #首先进行排序
res = []
i = 0
l = len(nums)
while i < l:
front = i+1
back = l - 1
target = - nums[i] #固定第一个元素并设定头尾两个flag
while front < back:
tmp = nums[front] + nums[back]
if tmp < target: #若过小则左flag右移
front += 1
elif tmp > target: #若过大则右flag左移
back -= 1
else: #正好则加入res
this = [nums[i], nums[front], nums[back]]
res += [this]
while front < back and nums[front] == this[1]: #左移移去重复
front += 1
while front < back and nums[back] == this[2]: #右移移去重复
back -= 1
while i < l-1 and nums[i+1] == nums[i]: #若遇到重复则后移
i+=1
i+=1
return res
将3Sum问题转化为2Sum问题
邻近的另一个问题3Sum Closest可以用类似方式进行解答
Given array nums = [-1, 2, 1, -4], and target = 1.
The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).
不重新上代码了，逻辑类似，sort后对i进行遍历，每次左右两个指针，故为N^2