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: 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]
]
题意：找到给定数组中的所有的和为0的三个数的组合，而且结果不允许重复。
思路：
首先将给定序列排序，用于后面的查找
每一组解有三个数，所以遍历给定序列的时候，设置三个索引：left,mid和right。
初始left=0,
对于 left from 0 to length-1:
mid=left+1,rigth=序列长度length-1
当mid<rigth时：
根据三者对应位置的值的和sum=arr[left]+arr[mid]+arr[rigth]来判断：
sum>0:则right左移
sum<0:则mid右移
sum==0:记录下来，right左移；mid右移
python代码实现：
1 class Solution(object): 2 def threeSum(self, nums): 3 """ 4 :type nums: List[int] 5 :rtype: List[List[int]] 6 """ 7 result = [] 8 left = 0 9 nums.sort() 10 length = len(nums) 11 while left < length - 2: 12 mid = left + 1 13 right = length - 1 14 while mid < right: 15 if nums[left] + nums[mid] + nums[right] > 0: 16 right -= 1 17 elif nums[left] + nums[mid] + nums[right] < 0: 18 mid += 1 19 else: 20 result.append([nums[left], nums[mid], nums[right]]) 21 mid += 1 22 right -= 1 23 while mid < right and nums[mid] == nums[mid - 1]: 24 mid += 1 25 while mid < right and nums[right] == nums[right + 1]: 26 right -= 1 27 left += 1 28 while left < length - 2 and nums[left] == nums[left - 1]: 29 left += 1 30 return result
提交结果：
原文：http://www.cnblogs.com/z941030/p/5608885.html