[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