标签：des style blog io ar color sp for on
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)
与3sum类似，逐渐递减到2sum。
1 vector<vector<int> > fourSum(vector<int> &num, int target) 2 { 3 int k = 0, l = 0; 4 int sum = 0; 5 vector<int> t(4); 6 vector<vector<int> > vt; 7 if (num.size() < 4) 8 return vt; 9 10 sort(num.begin(), num.end()); 11 for (int i = 0; i < num.size() - 3; i++) 12 { 13 if (i > 0 && num[i] == num[i - 1]) 14 continue; 15 for (int j = i + 1; j < num.size() - 2; j++) 16 { 17 if ((j > i + 1) && (num[j] == num[j - 1])) 18 continue; 19 k = j + 1; 20 l = num.size() - 1; 21 while (k < l) 22 { 23 if ((k > j + 1) && (num[k] == num[k - 1])) 24 { 25 k++; 26 continue; 27 } 28 if ((l < num.size() - 1) && (num[l] == num[l + 1])) 29 { 30 l--; 31 continue; 32 } 33 sum = num[i] + num[j] + num[k] + num[l]; 34 if (sum == target) 35 { 36 t.clear(); 37 t.push_back(num[i]); 38 t.push_back(num[j]); 39 t.push_back(num[k]); 40 t.push_back(num[l]); 41 vt.push_back(t); 42 k++; 43 l--; 44 } 45 else if (sum < target) 46 k++; 47 else 48 l--; 49 } 50 } 51 } 52 53 return vt; 54 }
时间复杂度：O(n^3)
标签：des style blog io ar color sp for on
原文地址：http://www.cnblogs.com/ym65536/p/4136148.html
