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]]
思路
暴力的话，时间复杂度是O(n^3）,试了下，肯定不行的，我开始想能不能转换2sum，就是先遍历一遍，求出两个数之和，然后再找这两个数之和存不存在就行。提交了一下，时间上还是过不了，想了半天，就只能用排序的方法，这样的话，我只需要从最左和最右开始找，有点二分的味道。这样的话，最坏情况是O(n^2)。
class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
length = len(nums)
nums.sort() #排序
last = length
result = []
for i in range(length - 2):
if(i > 0 and nums[i-1] == nums[i]):
continue #如果i和上次重复了，那么就不用找了！
l,r = i+1,length-1
flag = True
while(l < r):
s = nums[i] + nums[l] + nums[r]
if s < 0:
l += 1
elif s > 0:
r -= 1
else:
result.append([nums[i],nums[l],nums[r]])
while(l < r and nums[l] == nums[l+1]):
l += 1
while(l < r and nums[r] == nums[r-1]):
r -= 1
l += 1
r -= 1
return result