leetcode-15

15. 3Sum

Leetcode 15

I solved this problem using C++.
My idea: use two pointers to find the sum of three numbers.

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class Solution {
public:
    vector<vector<int>> threeSum(vector<int> &nums)
    {
        vector<vector<int>> result;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < nums.size(); i++)
        {
            // Skip duplicate elements
            if (i > 0 && nums[i] == nums[i - 1])
                continue;

            // set two pointers
            int lt = i + 1, rt = nums.size() - 1;
            int target = -nums[i];

            // set -nums[i] as target
            // find the sum of two numbers
            while (lt < rt)
            {
                int sum = nums[lt] + nums[rt];
                if (sum == target)
                {
                    result.push_back({nums[i], nums[lt], nums[rt]});
                    lt++;
                    rt--;
                    // Skip duplicate elements
                    while (lt < rt && nums[lt] == nums[lt - 1])
                        lt++;
                    while (lt < rt && nums[rt] == nums[rt + 1])
                        rt--;
                }
                else if (sum < target)
                {
                    lt++;
                }
                else
                {
                    rt--;
                }
            }
        }

        return result;
    }
};

e.g.
nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]

  1. sort the array
    nums = [-4,-1,-1,0,1,2]

  2. set i = 0, lt = 1, rt = 5

  3. set target = -nums[i] = 4
    ……

  4. set i = 1, lt = 2, rt = 5

  5. set target = -nums[i] = 1

  6. set sum = nums[lt] + nums[rt] = -1 + 2 = 1

  7. set result.push_back([-1, -1, 2])

  8. set lt = 3, rt = 4

  9. set sum = nums[lt] + nums[rt] = 0 + 1 = 1

  10. set result.push_back([-1, 0, 1])

  11. set lt = 4, rt = 3

  12. end of loop

conclusion

Only this method is easy to understand…😥