leetcode 18 四数之和 python

题目：给定一个包含 n 个整数的数组 nums 和一个目标值 target，判断 nums 中是否存在四个元素 a，b，c 和 d ，使得 a + b + c + d 的值与 target 相等？找出所有满足条件且不重复的四元组。

注意：

答案中不可以包含重复的四元组。

示例：

给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。

满足要求的四元组集合为：
做了两数之和，三数之和，现在是四数之和，有点吐血哈

[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

class Solution:
    def fourSum(self, nums: List[int], target: int) -> List[List[int]]:
    #简洁版
        def findNsum(nums, target, N, result, results):
            if len(nums) < N or N < 2 or target < nums[0]*N or target > nums[-1]*N:  # early termination
                return
            if N == 2: # two pointers solve sorted 2-sum problem
                l,r = 0,len(nums)-1
                while l < r:
                    s = nums[l] + nums[r]
                    if s == target:
                        results.append(result + [nums[l], nums[r]])
                        l += 1
                        while l < r and nums[l] == nums[l-1]:
                            l += 1
                    elif s < target:
                        l += 1
                    else:
                        r -= 1
            else: # recursively reduce N
                for i in range(len(nums)-N+1):
                    if i == 0 or (i > 0 and nums[i-1] != nums[i]):
                        findNsum(nums[i+1:], target-nums[i], N-1, result+[nums[i]], results)

        results = []
        findNsum(sorted(nums), target, 4, [], results)
        return results



        nums.sort()
        results = []
        self.findNsum(nums, target, 4, [], results)
        return results
#复杂版
#     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