【leetcode】3SUM python實現

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:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
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)

暴力法應該是O（N3） ，的時間複雜度，初步估計最少也需要O（N2） 的時間複雜度，所以先排序(O(Nlog(N))) ，從小到大，然後到正數的時候停下。
參考之前的twoSum，相當於是遍歷一遍，然後對每個遍歷的Si ，以−Si 爲target，往後找twoSum.
代碼如下：

class Solution(object):
    def threeSum(self, nums):
        nums = sorted(nums)
        res = []
        if len(nums)<3:
            return []
        for y in xrange(len(nums)):
            if nums[y]>0:
                break
            if y > 0 and nums[y] == nums[y-1]:
                continue
            res = self.twoSum(nums,-nums[y], y, res)
        return res

    def twoSum(self, nums, target, y, res):
        dict={}
        for i in xrange(len(nums)):
            if y >=i:
                continue
            x=nums[i]
            if target-x in dict:
                mlist = sorted([target-x, x, nums[y]])
                if mlist in res:
                    continue
                res.append(mlist)
                continue
            dict[x]=i
        return res