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Problem_18_FourSum.py
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83 lines (78 loc) · 3.43 KB
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class Solution(object):
'''
def fourSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[List[int]]
"""
'''
# 依然套用三数之和的双指针法, 前两个指针遍历, 后两个指针相向检索. 技巧是加入提前判定条件, 略过不必要的计算.
# O(N^3)
"""
def fourSum(self, nums, target):
nums.sort() # O(NlogN), 排序,便于从两头检索
result = []
for i in range(len(nums) - 3):
if i >= 1 and nums[i] == nums[i - 1]: # 跳过重复值
continue
if nums[0] + nums[1] + nums[2] + nums[3] > target: # 剪枝, 加速
break
if nums[i] + sum(nums[-3:]) < target: #剪枝,加速, 开头加末尾最大三个数,如果不满足,说明开头元素过小,直接跳过
continue
# 转为3sum
for j in range(i + 1, len(nums) - 2): # j 从 i的下一个数开始
if j > i + 1 and nums[j] == nums[j - 1]:
continue
if nums[i] + nums[j] + nums[-2] + nums[-1] < target: #开头加末尾最大两个数,如果不满足,说明开头元素过小,直接跳过
continue
if nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target: # 开头最小三个数都比target大,这组不满足,跳过
break
l = j + 1
r = len(nums) - 1
while l < r:
if nums[i] + nums[j] + nums[l] + nums[r] == target:
result.append([nums[i], nums[j], nums[l], nums[r]])
while l < r and nums[l] == nums[l + 1]: # 跳过相同的
l += 1
while l < r and nums[r] == nums[r - 1]: # 跳过相同的
r -= 1
l += 1
r -= 1
elif nums[i] + nums[j] + nums[l] + nums[r] < target: # 左边++,调大左值
l += 1
else: # 左边--,调小右值
r -= 1
return result
"""
# 采用递归的办法将N SUM削减到2 sum求解
def fourSum(self, nums, target):
nums.sort()
results = []
self.findNsum(nums, target, 4, [], results)
return results
def findNsum(self, nums, target, N, result, results):
if len(nums) < N or N < 2: return
# solve 2-sum
if N == 2:
l, r = 0, len(nums) - 1
while l < r:
if nums[l] + nums[r] == target:
results.append(result + [nums[l], nums[r]])
l += 1
r -= 1
while l < r and nums[l] == nums[l - 1]:
l += 1
while r > l and nums[r] == nums[r + 1]:
r -= 1
elif nums[l] + nums[r] < target:
l += 1
else:
r -= 1
else:
for i in range(0, len(nums) - N + 1): # careful about range
if target < nums[i] * N or target > nums[-1] * N: # take advantages of sorted list
break
if i == 0 or i > 0 and nums[i - 1] != nums[i]: # recursively reduce N
self.findNsum(nums[i + 1:], target - nums[i], N - 1, result + [nums[i]], results)
return