LeetCode[Array]----3Sum

3Sum

Given an array S of n integers, are there elements abc 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)

class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
res = []
if len(nums) < 3:
return []
# 对数组进行排序。方便后面的查找
nums.sort()
for i in range(len(nums) - 2):
# 把a作为target
target = nums[i] * -1
# 问题转化为b+c == -a的情况了
a = i + 1
b = len(nums) - 1
while a < b:
if nums[a] + nums[b] == target:
res.append([nums[i], nums[a], nums[b]])
a += 1
b -= 1
elif nums[a] + nums[b] > target:
b -= 1
else:
a += 1
dummyres = []
# 因为上面的方法产生的解有反复，须要去重
for i in range(len(res)):
if res[i] in dummyres:
pass
else:
dummyres.append(res[i])
return dummyres

class Solution(object):
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
res = []
if len(nums) < 3:
return []
nums.sort()
for i in range(len(nums) - 2):
# 当nums[i]大于0时。后面的元素和一定大于0，不须要再进行推断了
# 当nums[i]与前一个元素值同样时，解是同样的
if nums[i] > 0 or i and nums[i] == nums[i-1]:
continue
target = -nums[i]
left = i + 1
right = len(nums) - 1
while left < right:
# 当nums[right]小于0时。a,b,c三个元素值都小于0。也不须要进行-a == b+c的推断了
if nums[right] < 0:
break
if nums[left] + nums[right] == target:
res.append([nums[i], nums[left], nums[right]])
while left < right and nums[left+1] == nums[left]:  # 当nums[left]值与下一个元素值同样时，解同样
left += 1
while left < right and nums[right-1] == nums[right]:
right -= 1
left += 1
right -= 1
elif nums[left] + nums[right] > target:
right -= 1
else:
left += 1
return res

posted on 2017-08-15 18:48  blfbuaa  阅读(215)  评论(0编辑  收藏  举报