[Lintcode 3sum]三数之和（python，二分）

题目链接：http://www.lintcode.com/zh-cn/problem/3sum/?rand=true#

用这个OJ练练python…这个题意和解法就不多说了，O(n^2lgn)就行了，关键是！！python的语法…

要想给tuple排序，如果直接sort的话会自动转成list，这个时候要再转回来。

 class Solution:
     """
     @param numbersbers : Give an array numbersbers of n integer
     @return : Find all unique triplets in the array which gives the sum of zero.
     """
     def Solution(self):
         pass
     def threeSum(self, numbers):
         # write your code here
         numbers.sort()
         ret = []
         n = len(numbers)
         for i in range(0, n):
             for j in range(i+1, n):
                 d = -(numbers[j] + numbers[i])
                 l = 0
                 r = n - 1
                 p = -1
                 while l <= r:
                     m = (l + r) >> 1
                     if numbers[m] == d:
                         p = m
                         break
                     elif numbers[m] > d:
                         r = m - 1
                     elif numbers[m] < d:
                         l = m + 1
                 if p != -1 and p != i and p != j:
                     ret.append((numbers[i], numbers[j], numbers[p]))
         nn = len(ret)
         for i in range(0, nn):
             ret[i] = list(ret[i])
             ret[i].sort()
             ret[i] = tuple(ret[i])
         ret = list(set(ret))
         return ret