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)
4sum转为3sum,3sum转为2sum,代码如下:
1 class Solution { 2 public: 3 vector<vector<int> > fourSum(vector<int> &num, int target) { 4 sort(num.begin(), num.end()); //需要对数组元素进行排序 5 vector< vector<int> > ans; 6 for(int i=0; i+4<=num.size(); ++i) { 7 if( i>0 && num[i] == num[i-1] ) continue; 8 vector< vector<int> > threesum = threeSum(num, i+1, target-num[i]); 9 for(int j=0; j<threesum.size(); ++j) {10 threesum[j].insert(threesum[j].begin(), num[i]); //non-descending order11 ans.push_back(threesum[j]);12 }13 }14 return ans;15 }16 17 vector<vector<int> > threeSum(vector<int> &num, int left, int target) {18 vector< vector<int> > ans;19 for(int i=left; i+3<=num.size(); ++i) {20 if( i>left && num[i] == num[i-1] ) continue; //如果该数和前面数相同,那么需要直接跳过21 vector< vector<int> > twoans = twoSum(num, i+1, num.size()-1, target-num[i]); //直接转为twosum问题22 for(int j=0; j<twoans.size(); ++j) {23 twoans[j].insert(twoans[j].begin(), num[i]); //non-descending order24 ans.push_back(twoans[j]);25 }26 }27 return ans;28 }29 30 vector< vector<int> > twoSum(vector<int>& num, int left, int right, int target) {31 vector< vector<int> > ans;32 vector<int> v;33 while( left < right ) {34 int sum = num[left] + num[right];35 if( sum == target ) { //找到一个解36 v.push_back(num[left]);37 v.push_back(num[right]);38 ans.push_back(v);39 v.clear();40 ++left; //left要后移至不同于前一个数位置上41 while( left < right && num[left] == num[left-1] ) ++left;42 --right; //right要前移至不同于后一个数位置上43 while( left < right && num[right] == num[right+1] ) --right;44 }45 else if( sum > target ) --right;46 else ++left;47 }48 return ans;49 }50 };