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：
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
result = []
left = 0
nums.sort()
length = len(nums)
while left < length - 2:
mid = left + 1
right = length - 1
while mid < right:
if nums[left] + nums[mid] + nums[right] > 0:
right -= 1
elif nums[left] + nums[mid] + nums[right] < 0:
mid += 1
else:
result.append([nums[left], nums[mid], nums[right]])
mid += 1
right -= 1
while mid < right and nums[mid] == nums[mid - 1]:
mid += 1
while mid < right and nums[right] == nums[right + 1]:
right -= 1
left += 1
while left < length - 2 and nums[left] == nums[left - 1]:
left += 1
return result
