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Problem Definition:
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:
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)
Solution:
1）最直接的思路，固定第一个数，变成 3 Sum 问题，进而变成2 Sum 问题；这个想法可以扩展到k-Sum。重复元组的问题通过跳过重复元素来避免。
1 # @param {integer[]} nums 2 # @param {integer} target 3 # @return {integer[][]} 4 def fourSum(nums, target): 5 nums.sort() 6 res=[] 7 first=0 8 finalFirst=len(nums)-4 9 finalSecond=len(nums)-3 10 rightBound=len(nums)-1 11 while first<=finalFirst: 12 nf=nums[first] 13 second=first+1 14 while second<=finalSecond: 15 left=second+1 16 right=rightBound 17 ns=nums[second] 18 while left<right: 19 nl=nums[left] 20 nr=nums[right] 21 four=nf+ns+nl+nr 22 if four>target: 23 right-=1 24 elif four<target: 25 left+=1 26 else: 27 res+=[nf, ns, nl, nr], 28 while left<right and nums[left]==nl: 29 left+=1 30 while left<right and nums[right]==nr: 31 right-=1 32 second+=1 33 while second<=finalSecond and nums[second]==ns: 34 second+=1 35 first+=1 36 while first<=finalFirst and nums[first]==nf: 37 first+=1 38 return res
2）HashTable. 数组中元素两两求和，以和为key，对应的value是多个四元组组成的数组，每个四元组保存了和为key的两个元素，及他们的下标。O(n^2)时间形成这个表。
然后遍历HashTable [O(n)]，对key1，找到能够相加得target的key2 [O(1)]，把这俩key对应的value内的四元组交叉组合。防止重复是关键。可以用Set（集合）来存储这些交叉后的元组。
1 # @param {integer[]} nums 2 # @param {integer} target 3 # @return {integer[][]} 4 def fourSum(self, nums, target): 5 n=len(nums) 6 if n<4: 7 return [] 8 vsMap={} 9 res=[] 10 for i in range(n): 11 for j in range(i+1,n): 12 key=nums[i]+nums[j] 13 vsMap.setdefault(key, []) 14 vsMap[key].append([nums[i], i, nums[j], j]) 15 for key in vsMap: 16 need=target-key 17 if need in vsMap: 18 pB=vsMap[need] 19 else: 20 continue 21 pA=vsMap[key] 22 for a in pA: 23 for b in pB: 24 if a[1]==b[1] or a[3]==b[3] or a[1]==b[3] or a[3]==b[1]: 25 continue 26 cmb=sorted([a[0], a[2], b[0], b[2]]) 27 if cmb not in res: 28 res+=cmb, 29 return res
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原文地址：http://www.cnblogs.com/acetseng/p/4694587.html
