题目:原题链接(中等)
标签:数组、哈希表、双指针
解法 | 时间复杂度 | 空间复杂度 | 执行用时 |
Ans 1 (Python) | O ( N 4 ) | O ( 1 ) | 80ms (94.42%) |
Ans 2 (Python) | |||
Ans 3 (Python) |
解法一:
class Solution:
def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
# 处理特殊情况
if not nums or len(nums) < 4:
return []
ans = []
N = len(nums)
nums.sort()
for i1 in range(N - 3):
if i1 > 0 and nums[i1] == nums[i1 - 1]:
continue
if nums[i1] + nums[i1 + 1] + nums[i1 + 2] + nums[i1 + 3] > target:
break
if nums[i1] + nums[-3] + nums[-2] + nums[-1] < target:
continue
for i2 in range(i1 + 1, N - 2):
if i2 > i1 + 1 and nums[i2] == nums[i2 - 1]:
continue
if nums[i1] + nums[i2] + nums[i2 + 1] + nums[i2 + 2] > target:
break
if nums[i1] + nums[i2] + nums[-2] + nums[-1] < target:
continue
i3, i4 = i2 + 1, N - 1
while i3 < i4:
val = nums[i1] + nums[i2] + nums[i3] + nums[i4]
if val == target:
# print(i1, i2, i3, i4, "->", nums[i1], nums[i2], nums[i3], nums[i4])
ans.append([nums[i1], nums[i2], nums[i3], nums[i4]])
while i3 < i4 and nums[i3] == nums[i3 + 1]:
i3 += 1
i3 += 1
while i3 < i4 and nums[i4] == nums[i4 - 1]:
i4 -= 1
i4 -= 1
elif val < target:
i3 += 1
else:
i4 -= 1
return ans