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: