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題目:
給你一個包含 n 個整數的數組nums
,判斷nums
中是否存在三個元素a,b,c
,使得a + b + c = 0
?請你找出所有滿足條件且不重複的三元組。
注意:答案中不可以包含重複的三元組。
示例:
- 示例 1:
給定數組 nums = [-1, 0, 1, 2, -1, -4], 滿足要求的三元組集合爲: [ [-1, 0, 1], [-1, -1, 2] ]
代碼
方法一: 集合的思路
執行用時:676 ms, 在所有 Python3 提交中擊敗了93.05%的用戶
內存消耗:17 MB, 在所有 Python3 提交中擊敗了6.02%的用戶
class Solution:
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
res = []
nums.sort()
for i in range(len(nums)):
if nums[i] > 0:
break
if i >= 1 and nums[i] == nums[i-1]:
continue
target = -nums[i]
cache = set()
for j in range(i+1, len(nums)):
if nums[j] in cache:
if len(res) == 0 or res[-1]!=[nums[i], target-nums[j], nums[j]]:
res.append([nums[i], target-nums[j], nums[j]])
cache.add(target-nums[j])
return res
"""
For Example: input: nums = [-1, 0, 1, 2, -1, -4]
output: [
[-1, 0, 1],
[-1, -1, 2]
]
"""
nums = [-1, 0, 1, 2, -1, -4]
solution = Solution()
result = solution.threeSum(nums)
print('輸出爲:', result)
方法二: 排序+雙指針
執行用時 :44 ms, 在所有 Python3 提交中擊敗了59.87%的用戶
內存消耗 :13.7 MB, 在所有 Python3 提交中擊敗了6.15%的用戶
class Solution:
def threeSum(self, nums):
"""
:type nums: List[int]
:rtype: List[List[int]]
"""
n=len(nums)
res=[]
if(not nums or n<3):
return []
nums.sort()
res=[]
for i in range(n):
if(nums[i]>0):
return res
if(i>0 and nums[i]==nums[i-1]):
continue
L=i+1
R=n-1
while(L<R):
if(nums[i]+nums[L]+nums[R]==0):
res.append([nums[i],nums[L],nums[R]])
while(L<R and nums[L]==nums[L+1]):
L=L+1
while(L<R and nums[R]==nums[R-1]):
R=R-1
L=L+1
R=R-1
elif(nums[i]+nums[L]+nums[R]>0):
R=R-1
else:
L=L+1
return res
"""
For Example: input: nums = [-1, 0, 1, 2, -1, -4]
output: [
[-1, 0, 1],
[-1, -1, 2]
]
"""
nums = [-1, 0, 1, 2, -1, -4]
solution = Solution()
result = solution.threeSum(nums)
print('輸出爲:', result)
參考
- https://www.bilibili.com/video/BV1ek4y1d7sw
- https://leetcode-cn.com/problems/3sum/solution/pai-xu-shuang-zhi-zhen-zhu-xing-jie-shi-python3-by/