15. 3Sum

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)

AC代码：

class Solution(object):
    def threeSum(self, nums):
        ret_list = []
        nums.sort()
        for i, vi in enumerate(nums[:-2]):
            if i - 1 >= 0 and vi == nums[i - 1]: continue
            temp_list = []
            j = i + 1
            while j < len(nums):
                need_val = - vi - nums[j];
                if need_val in temp_list:
                    ret_list.append([vi, need_val, nums[j]])
                    temp_list.remove(need_val)
                elif nums[j] not in nums[i+1: j]:
                    temp_list.append(nums[j])
                j += 1
        return ret_list

此方法较为非主流，核心思想是用一个临时列表用来存放遍历过不符合要求的值，每次取新值来运算，如果和等于0则这一条加入返回列表。为了防止重复还需要在临时列表中删除已经合法的值。效率一般。

较为主流的解法是用两个指针：

class Solution(object):
    def threeSum(self, nums):
        ret_list = []
        nums.sort()
        for i_1st in xrange(len(nums) - 2):
            if nums[i_1st] > 0: break
            # remove duplicates
            if i_1st > 0 and nums[i_1st] == nums[i_1st - 1]: continue
            i_2nd, i_3rd = i_1st + 1, len(nums) - 1
            while i_2nd < i_3rd:
                s = nums[i_1st] + nums[i_2nd] + nums[i_3rd]
                if s > 0:
                    i_3rd -= 1
                elif s < 0:
                    i_2nd += 1
                else:
                    ret_list.append([nums[i_1st], nums[i_2nd], nums[i_3rd]])
                    # remove duplicates
                    while i_2nd < i_3rd and nums[i_2nd] == nums[i_2nd + 1]:
                        i_2nd += 1
                    while i_2nd < i_3rd and nums[i_3rd] == nums[i_3rd - 1]:
                        i_3rd -= 1
                    i_2nd += 1; i_3rd -= 1
        return ret_list

因为首先进行了排序，所以用两个指针分别从首尾两端进行缩小，效率更高。注意处理重复的情况。

上面两种方法都需要先进行排序。