# 算法---求四数之和

``````#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
__title__ = ''
__author__ = 'thinkreed'
__mtime__ = '2017/3/19'

idea from https://discuss.leetcode.com/topic/22705/python-140ms-beats-100-and-works-for-n-sum-n-2/4
"""

class Solution(object):
def fourSum(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: List[List[int]]
"""

def find_N_sum(nums, start, target, n, result, results):
#跳过剩余数组长度小于n或者当前的维次2或者target小于第一个数而太小不可能存在或者太大不存在的情况
if len(nums[start:]) < n or n < 2 or target < nums[start] * n or target > nums[-1] * n:
return
#求两数和为target
elif n == 2:
left, right = start, len(nums) - 1

while left < right:
cur_sum = nums[left] + nums[right]

if cur_sum < target:
left += 1
elif cur_sum > target:
right -= 1
else:
results.append(result + [nums[left], nums[right]])
left += 1
#跳过重复
while left < right and nums[left] == nums[left - 1]:
left += 1

#降维次，将求n数和变为求n-1数和
else:
for i in range(len(nums[start:]) - n + 1):
if i == 0 or (i > 0 and nums[start + i - 1] != nums[start + i]):
find_N_sum(nums, start + i + 1, target - nums[start + i], n - 1, result + [nums[start + i]],
results)

results = []
find_N_sum(sorted(nums), 0, target, 4, [], results)
return results

if __name__ == '__main__':
print(Solution().fourSum([1, 0, -1, 0, -2, 2], 0))
``````