LeetCode: 3Sum

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：

class Solution {
 public:
     vector<vector<int> > threeSum(vector<int> &num) {
         // Start typing your C/C++ solution below
         // DO NOT write int main() function
         sort(num.begin(), num.end());
         vector<vector<int>> ret;
         vector<int> tmp;
         if (num.size() < 3) return ret;
         for (int i = 0; num[i] <= 0 && i < num.size()-2; i++) {
             int end = num.size()-1;
             if (i > 0 && num[i] == num[i-1]) continue;
             for(int beg = i+1; beg < end; beg++) {
                 if (beg > i+1 && num[beg] == num[beg-1]) continue;
                 int sum = num[i] + num[beg] + num[end];
                 if (sum == 0) {
                     tmp.clear();
                     tmp.push_back(num[i]);
                     tmp.push_back(num[beg]);
                     tmp.push_back(num[end]);
                     ret.push_back(tmp);
                 }
                 else if (sum > 0) end--, beg--;
             }
         }
         return ret;
     }
 };

C#代码：

public class Solution {
    public IList<IList<int>> ThreeSum(int[] nums) {
        IList<IList<int>> ans = new List<IList<int>>();
        Array.Sort(nums);
        if (nums.Count() < 3) return ans;
        for (int i = 0; i < nums.Count()-2; i++) {
            int end = nums.Count() - 1;
            if (i > 0 && nums[i] == nums[i-1]) continue;
            for (int beg = i + 1; beg < end; beg++) {
                if (beg > i + 1 && nums[beg] == nums[beg-1]) continue;
                int sum = nums[i] + nums[beg] + nums[end];
                if (sum == 0) {
                    List<int> tmp = new List<int>(){nums[i], nums[beg], nums[end]};
                    ans.Add(tmp);
                }
                else if (sum > 0) {
                    end--;
                    beg--;
                }
            }
        }
        return ans;
    }
}

 Java:

public class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> ans = new ArrayList<List<Integer>>();
        Arrays.sort(nums);
        for (int i = 0; i < nums.length; i++)
        {
            if (i > 0 && nums[i] == nums[i-1])
                continue;
            int left = i + 1;
            int right = nums.length - 1;
            while (left < right)
            {
                if (left > i+1 && nums[left] == nums[left-1]) {
                    left++;
                    continue;
                }
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0) {
                    List<Integer> tmp = new ArrayList<Integer>();
                    tmp.add(nums[i]);
                    tmp.add(nums[left]);
                    tmp.add(nums[right]);
                    ans.add(tmp);
                    left++;
                }
                else if (sum > 0) right--;
                else left++;
            }
        }
        return ans;
    }
}