原题链接：https://leetcode.com/problems/4sum/

思路：类似于之前的3sum的解法，3sum采用了固定一个元素，另两个元素从两面夹逼；4sum则两个元素从左往右遍历，另两个元素夹逼。注意中间需要判断数字是否相同，若相同则一直加1

代码：

class Solution(object):
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        if len(nums)<4:
            return []

        result = []
        nums = sorted(nums)
        i = 0

        while i < len(nums) - 3:
            j = i + 1
            sum3 = target - nums[i]
            while j < len(nums) - 2:
                sum2 = sum3 - nums[j]
                m = j + 1
                n = len(nums) - 1
                while m < n:
                    if nums[m] + nums[n] > sum2:
                        while n > m:
                            n -= 1
                            if nums[n] != nums[n+1]:
                                break
                    elif nums[m] + nums[n] < sum2:
                        while n > m:
                            m += 1
                            if nums[m] != nums[m-1]:
                                break
                    else:
                        result.append([nums[i], nums[j], nums[m], nums[n]])
                        while n > m:
                            m += 1
                            if nums[m] != nums[m-1]:
                                break
                        while n > m:
                            n -= 1
                            if nums[n] != nums[n+1]:
                                break
                while j < len(nums)-2:
                    j += 1
                    if nums[j] != nums[j-1]:
                        break
            while i < len(nums)-3:
                i += 1
                if nums[i] != nums[i-1]:
                    break

        return result

参考代码：https://blog.csdn.net/qq_28119401/article/details/52972606
在这里再记录另外一种方法，下次自己再尝试一下：
https://www.cnblogs.com/zuoyuan/p/3699384.html