题目：
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] ]
题解：
这个题与3Sum类似，求4Sum就在原基础上再加上一层循环就可以了。这里只给出此种解法思路的其中一个解法。
Solution 1 (36ms)
1 class Solution { 2 public: 3 vector<vector<int>> fourSum(vector<int>& nums, int target) { 4 set<vector<int>> sv; 5 sort(nums.begin(), nums.end()); 6 int n = nums.size(); 7 8 for(int i=0; i<n-3; i++) { 9 for(int j=i+1; j<n-2; j++) { 10 int k = j+1, l = n-1; 11 int a = nums[i], b = nums[j]; 12 while(k<l) { 13 int c = nums[k], d = nums[l]; 14 if(a+b+c+d == target) { 15 sv.insert({a,b,c,d}); 16 k++; 17 l--; 18 } 19 else if(a+b+c+d < target) k++; 20 else l--; 21 } 22 } 23 } 24 return vector<vector<int>> (sv.begin(),sv.end()); 25 } 26 };
还有一种解法，也是利用了3Sum，不过不是再加一层循环，而是直接调用3Sum函数：取nums[i]，然后对后续剩余数组元素求3Sum，tar为target - nums[i]；
Solution 2 (32ms)
1 class Solution { 2 public: 3 vector<vector<int> > threeSum(vector<int> &nums, int target) { 4 set<vector<int>> sv; 5 sort(nums.begin(), nums.end()); 6 int n = nums.size(); 7 8 for(int i=0; i<n-2; i++) { 9 int a = nums[i]; 10 int j = i+1, k = n-1; 11 while(j<k) { 12 int b = nums[j], c = nums[k]; 13 if(a+b+c == target) { 14 sv.insert({a,b,c}); 15 j++; 16 k--; 17 } 18 else if(a+b+c > target) k--; 19 else j++; 20 } 21 } 22 return vector<vector<int>> (sv.begin(),sv.end()); 23 } 24 vector<vector<int>> fourSum(vector<int>& nums, int target) { 25 vector<vector<int>> vv; 26 sort(nums.begin(), nums.end()); 27 int n = nums.size(); 28 29 for(int i=0; i<n-3; i++) { 30 if(i>0 && nums[i] == nums[i-1]) continue; 31 //截取剩余数组 32 vector<int> v(nums.begin()+i+1,nums.end()); 33 vector<vector<int>> tmp = threeSum(v, target - nums[i]); 34 for(int j=0; j<tmp.size(); j++) { 35 tmp[j].insert(tmp[j].begin(), nums[i]); 36 vv.push_back(tmp[j]); 37 } 38 } 39 return vv; 40 } 41 };
还有一种更为优化的解法，思路是一致的，只是加了一个小技巧：在两个外循环中先判断四个值的和与target的大小，即
if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break;
if(nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target) continue;
通过这两条语句减少了搜索时间，不必进入内循环判断；对于j同理。
Solution 3 (12ms)
1 class Solution { 2 public: 3 vector<vector<int>> fourSum(vector<int>& nums, int target) { 4 set<vector<int>> sv; 5 sort(nums.begin(), nums.end()); 6 int n = nums.size(); 7 if(n<4) return vector<vector<int>> (sv.begin(),sv.end()); 8 for(int i=0; i<n-3; i++) { 9 if(nums[i]+nums[i+1]+nums[i+2]+nums[i+3]>target) break; 10 if(nums[i]+nums[n-3]+nums[n-2]+nums[n-1]<target) continue; 11 if(i>0&&nums[i]==nums[i-1]) continue; 12 for(int j=i+1; j<n-2; j++) { 13 if(j>i+1&&nums[j]==nums[j-1]) continue; 14 if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target) break; 15 if(nums[i]+nums[j]+nums[n-2]+nums[n-1]<target) continue; 16 int k = j+1, l = n-1; 17 int a = nums[i], b = nums[j]; 18 while(k<l) { 19 int c = nums[k], d = nums[l]; 20 if(a+b+c+d == target) { 21 sv.insert({a,b,c,d}); 22 k++; 23 l--; 24 } 25 else if(a+b+c+d < target) k++; 26 else l--; 27 } 28 } 29 } 30 return vector<vector<int>> (sv.begin(),sv.end()); 31 } 32 };
Solution 4
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { int n = nums.size(); vector<vector<int>> res; sort(nums.begin(), nums.end()); for(int i = 0; i < n - 3; ++i){ for(int j = i + 1; j < n - 2; ++j){ int begin = j + 1, end = n - 1; while(begin < end){ int sum = nums[i] + nums[j] + nums[begin] + nums[end]; if(sum == target) res.push_back({nums[i], nums[j], nums[begin++], nums[end--]}); else if (sum < target) ++begin; else --end; } } } sort(res.begin(), res.end()); res.erase(unique(res.begin(), res.end()), res.end()); return res; } };
去重的另一种方法， 使用了algorithm
Solution 5
class Solution { public: vector<vector<int>> fourSum(vector<int>& nums, int target) { int n = nums.size(); vector<vector<int>> res; sort(nums.begin(), nums.end()); unordered_multimap<int, pair<int, int>> map; for(int i = 0; i < n - 1; ++i){ for(int j = i + 1; j < n; ++j){ map.insert(make_pair(nums[i] + nums[j], make_pair(i, j))); } } for(auto i = map.begin(); i != map.end(); ++i){ int val = target - i->first; auto range = map.equal_range(val); for(auto j = range.first; j != range.second; ++j){ auto a = i->second.first, b = i->second.second; auto c = j->second.first, d = j->second.second; if(a != c && a != d && b != c && b != d){ vector<int> tmp = {nums[a], nums[b], nums[c], nums[d]}; sort(tmp.begin(), tmp.end()); res.push_back(tmp); } } } sort(res.begin(), res.end()); res.erase(unique(res.begin(), res.end()), res.end()); return res; } };
先缓存两个数的和，注意要使用multimap（from 九章算法）