这一题的关键是不能用O(N^3)的算法,肯定会LTE,O(N^2)的算法在于里面一层用贪心的方法向中间逼近
逼近的速度可以用二分法优化,但是鉴于比较复杂这里就没写了
C++代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 vector<vector<int> > threeSum(vector<int> &num) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 sort(num.begin(), num.end()); 7 vector<vector<int>> ret; 8 vector<int> tmp; 9 if (num.size() < 3) return ret; 10 for (int i = 0; num[i] <= 0 && i < num.size()-2; i++) { 11 int end = num.size()-1; 12 if (i > 0 && num[i] == num[i-1]) continue; 13 for(int beg = i+1; beg < end; beg++) { 14 if (beg > i+1 && num[beg] == num[beg-1]) continue; 15 int sum = num[i] + num[beg] + num[end]; 16 if (sum == 0) { 17 tmp.clear(); 18 tmp.push_back(num[i]); 19 tmp.push_back(num[beg]); 20 tmp.push_back(num[end]); 21 ret.push_back(tmp); 22 } 23 else if (sum > 0) end--, beg--; 24 } 25 } 26 return ret; 27 } 28 };
C#代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 public IList<IList<int>> ThreeSum(int[] nums) { 3 IList<IList<int>> ans = new List<IList<int>>(); 4 Array.Sort(nums); 5 if (nums.Count() < 3) return ans; 6 for (int i = 0; i < nums.Count()-2; i++) { 7 int end = nums.Count() - 1; 8 if (i > 0 && nums[i] == nums[i-1]) continue; 9 for (int beg = i + 1; beg < end; beg++) { 10 if (beg > i + 1 && nums[beg] == nums[beg-1]) continue; 11 int sum = nums[i] + nums[beg] + nums[end]; 12 if (sum == 0) { 13 List<int> tmp = new List<int>(){nums[i], nums[beg], nums[end]}; 14 ans.Add(tmp); 15 } 16 else if (sum > 0) { 17 end--; 18 beg--; 19 } 20 } 21 } 22 return ans; 23 } 24 }
Java:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 public List<List<Integer>> threeSum(int[] nums) { 3 List<List<Integer>> ans = new ArrayList<List<Integer>>(); 4 Arrays.sort(nums); 5 for (int i = 0; i < nums.length; i++) 6 { 7 if (i > 0 && nums[i] == nums[i-1]) 8 continue; 9 int left = i + 1; 10 int right = nums.length - 1; 11 while (left < right) 12 { 13 if (left > i+1 && nums[left] == nums[left-1]) { 14 left++; 15 continue; 16 } 17 int sum = nums[i] + nums[left] + nums[right]; 18 if (sum == 0) { 19 List<Integer> tmp = new ArrayList<Integer>(); 20 tmp.add(nums[i]); 21 tmp.add(nums[left]); 22 tmp.add(nums[right]); 23 ans.add(tmp); 24 left++; 25 } 26 else if (sum > 0) right--; 27 else left++; 28 } 29 } 30 return ans; 31 } 32 }