python 三数之和 多种解法
原创
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方法一：暴力法
def threeSum(nums):
result = []
nums.sort()
for i in range(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
for j in range(i+1, len(nums)-1):
if j > i+1 and nums[j] == nums[j-1]:
continue
for k in range(j+1, len(nums)):
if k > j+1 and nums[k] == nums[k-1]:
continue
if nums[i] + nums[j] + nums[k] == 0:
result.append([nums[i], nums[j], nums[k]])
return result
方法二：双指针法
def threeSum(nums):
result = []
nums.sort()
for i in range(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
left, right = i+1, len(nums)-1
while left < right:
total = nums[i] + nums[left] + nums[right]
if total < 0:
left += 1
elif total > 0:
right -= 1
else:
result.append([nums[i], nums[left], nums[right]])
while left < right and nums[left] == nums[left+1]:
left += 1
while left < right and nums[right] == nums[right-1]:
right -= 1
left += 1
right -= 1
return result
方法三：哈希表法
def threeSum(nums):
result = []
nums.sort()
for i in range(len(nums)-2):
if i > 0 and nums[i] == nums[i-1]:
continue
seen = set()
j = i + 1
while j < len(nums):
complement = -nums[i] - nums[j]
if complement in seen:
result.append([nums[i], complement, nums[j]])
while j + 1 < len(nums) and nums[j] == nums[j+1]:
j += 1
seen.add(nums[j])
j += 1
return result
