leetcode 18 4Sum

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:

• 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)

类似于leetcode 15 3sum。采用类似的思路，O(n^3)，超时。。

class Solution:    def find(self,s,start,end,num):        while start<=end:            mid=(start+end)/2            if s[mid]==num:                return True            elif s[mid]>num:                end=mid-1            else:                start=mid+1        return False    # @return a list of lists of length 3, [[val1,val2,val3]]    def fourSum(self, num, target):        a=sorted(num)        l=len(a)        lst=[]        for i in range(l-3):            if i>=1 and a[i-1]==a[i]:                continue            for j in range(i+1,l-2):                if j!=i+1 and a[j]==a[j-1]:                    continue                for k in range(j+1,l-1):                    if k!=j+1 and a[k]==a[k-1]:                        continue                    if self.find(a,k+1,l-1,target-a[i]-a[j]-a[k]):                        lst.append( [a[i],a[j],a[k],target-a[i]-a[j]-a[k]] )        return lst

在此基础上优化：

增加一个字典缓存exist={}，把每两个数的和都加入字典；如果target-a[i]-a[j]不在字典中，则不必继续查找。时间复杂度平均降到O(n^2)

    def fourSum(self, num, target):        a=sorted(num)        l=len(a)        lst=[]        exist={}       <span style="color:#ff6666;"> for i in range(l-1):            for j in range(l):                exist[a[i]+a[j]]=True</span>        for i in range(l-3):            if i>=1 and a[i-1]==a[i]:                continue            for j in range(i+1,l-2):                if j!=i+1 and a[j]==a[j-1]:                    continue                <span style="color:#ff0000;">if exist.get(target-a[i]-a[j],False):</span>                    for k in range(j+1,l-1):                        if k!=j+1 and a[k]==a[k-1]:                            continue                        if self.find(a,k+1,l-1,target-a[i]-a[j]-a[k]):                            lst.append( [a[i],a[j],a[k],target-a[i]-a[j]-a[k]] )        return lst