15. 3Sum hash解法 vs 雙指標解法
阿新 • • 發佈:2019-01-08
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note: The solution set must not contain duplicate triplets.
For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
hash解法 120ms AC
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { sort(nums.begin(), nums.end()); vector<vector<int>> res; for(int i = 0; i < nums.size(); i++){ if(i > 0 && nums[i] == nums[i - 1]) continue; vector<vector<int>> two = twoSum(nums, i + 1, 0 - nums[i]); if(two.size() == 0) continue; else{ for(auto it : two){ it.push_back(nums[i]); res.push_back(it); } } } return res; } vector<vector<int>> twoSum(vector<int>& nums, int start, int target){ int n = nums.size(); vector<vector<int>> res; if(start >= n - 1) return res; unordered_map<int, int> ht; for(int i = start; i < n; i++){ vector<int> temp; if(i < n - 1 && nums[i] == nums[i + 1] && 2 * nums[i] == target ){ temp = {nums[i], nums[i]}; res.push_back(temp); while(i <= n - 2 && nums[i] == nums[i + 1]) i++; }else{ if(i > start && nums[i] == nums[i - 1]) continue; int tofind = target - nums[i]; if(ht.find(tofind) != ht.end() ){ temp = {nums[i], nums[ht[tofind]]}; res.push_back(temp); }//else 注意這個else千萬不能要 ht[nums[i]] = i; } } return res; } };
雙指標解法:
class Solution { public: vector<vector<int>> threeSum(vector<int>& nums) { int n = nums.size(); sort(nums.begin(), nums.end()); vector<vector<int>> res; for(int i = 0; i < n; i ++){ if(i > 0 && nums[i] == nums[i - 1]) continue; int L = i + 1, R = n - 1; vector<vector<int>> two; vector<int> temp; int target = 0 - nums[i]; if(target < 0)//妙哉! { break; } while(L < R){ if(nums[L] + nums[R] < target){ L++; }else if(nums[L] + nums[R] > target){ R--; }else{ /*vector<int> triplet(3, 0); triplet[0] = num[i]; triplet[1] = num[front]; triplet[2] = num[back]; res.push_back(triplet);*///這樣寫更佳,會省一些程式碼和for(auto it : two)的時間 temp = {nums[L], nums[R]}; two.push_back(temp); L++; R--; } while (L > i + 1 && L < n && nums[L] == nums[L - 1]) L++; while (R < n - 1 && R > i && nums[R] == nums[R + 1]) R--; } if(two.size() == 0) continue; else{ for(auto it : two){ it.push_back(nums[i]); res.push_back(it); } } } return res; } };