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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)
思路：
土一点的解法就是一个一个判断，让四个加起来的和等于target。。然而这样太土了，python表示又TLE啦～～～
class Solution(object): def fourSum(self, nums, target): ans=[] if len(nums)>=4: for i in range(len(nums)-4): for j in range(i+1,len(nums)): for k in range(j+1,len(nums)): for l in range(k+1,len(nums)): if nums[i]+nums[j]+nums[k]+nums[l]==target: temp=sorted([nums[i],nums[j],nums[k],nums[l]]) if temp not in ans: ans.append(temp) return ans
好看一点的解法就是，两两求和，存在一个字典里面，key=两数之和，其value存储了该数的对应下标。
用到的函数 enumeration()可以遍历数组，以及返回其下标
for i in enumerate([5,7]): print i(0,5)（1,7）
itertools.combinations( XX ,2)函数
随机取两个数且不重复 即ab,ba视为aba=isdisjoin(b):a元素和b元素的交集为空集，则返回为True
import collectionsimport itertoolsclass Solution: def fourSum(self,nums,target): doc=collections.defaultdict(list) ans=[] for (id1,val1),(id2,val2) in itertools.combinations(enumerate(nums),2): doc[val1+val2].append({id1,id2}) keys=doc.keys() for key in keys: if doc[key] and doc[target-key]: for part1 in doc[key]: for part2 in doc[target-key]: if part1.isdisjoint(part2): temp=sorted([nums[i] for i in part1 | part2]) if temp not in ans: ans.append(temp) del doc[key] if key!=target-key: del doc[target-key] return ans 转载地址：http://twqmi.baihongyu.com/