18. 4Sum Leetcode Python

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:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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)

这题的解法和前面3sum的区别在于如果采用之前的解法，时间复杂度变为O(n^3) 因此我们考虑采用hash tale来用空间换时间。 基本做法是先把两两的和作为key存在hashtable,对应的 value是两两值的index.

然后再遍历时候用target -num[i]-num[j] 如果得到的差值还在hashtable里面就将其组合成所要的解。 时间复杂度为O(n^2) 空间复杂度为O (n^2)（这里我还在考虑因为是两两选出来，所以应该有C(n,2)个）

We can trade space with time in this problem. first iterate the num and use 2 sum as the key ,indexes as the value in hastable.

Then we start iterate again. take target -num[i]-num[j]，if it is still in the hashtable, we take out the values and make the solution. The overall time and space complexity are both O(n^2)

code:

class Solution:
    # @return a list of lists of length 4, [[val1,val2,val3,val4]]
    def fourSum(self, num, target):
        solution=[]
        num.sort()
        dict={}
        if len(num)<4:
            return solution
        for i in range(len(num)):
            #if i>0 and num[i]==num[i+1]:
             #   continue
            for j in range(i+1,len(num)):
                val=num[i]+num[j]
                if val not in dict:
                    dict[val]=[[i,j]]
                else:
                    dict[val].append([i,j])
        for i in range(len(num)):
           # if i>0 and num[i]==num[i+1]:## we should delete this here
            #    continue
            for j in range(i+1,len(num)-2):
                dif=target-num[i]-num[j]
                if dif in dict:
                    for k in dict[dif]:
                        if k[0]>j and [num[i],num[j],num[k[0]],num[k[1]]] not in solution:
                            solution.append([num[i],num[j],num[k[0]],num[k[1]]])
        return solution