4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

这题有多种解法，使用two pointer的解法复杂度是O(n^3)的，枚举前两位，后两位是双指针，这种解法很trival。另外一种解法是使用hashmap先保存两两的pair。

然后采用two sum的解法。值得注意的是two sum那题是数组里所有的数字都不一样，但是这里不一样，pair很多的和值是一样的，数字本身也可能有重复，所以去重非常关键。代码如下：

class Solution(object):
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        if not nums or len(nums) < 4:
            return []
        nums.sort()
        res = set()
        map = {}
        for i in xrange(0,len(nums)):
            for j in xrange(i+1, len(nums)):
                key = nums[i] + nums[j]
                if key not in map:
                     map[key] = [(nums[i],nums[j],i,j)]
                else:
                     map[key].append((nums[i],nums[j],i,j))
        for key, value in map.iteritems():
            sub = target - key
            if key <= target/2 and sub in map:
                for a in value:
                    for b in map[sub]:
                        if a[2] != b[2] and a[3]!= b[3] and a[3] != b[2] and a[2]!=b[3]:
                            res.add(tuple(sorted([a[0],b[0],a[1],b[1]])))
        return [list(a) for a in res]

这题个人觉得最坏的时间复杂度是O(n^4)的，数组全部都是target/4的情况，具体还需要再思考，详见一个leetcode讨论和另外一个一亩三分地的帖子,一个很好的解法http://www.cnblogs.com/TenosDoIt/p/3649607.html