【leetcode】4Sum（middle）

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:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

 

    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)

 

思路：

讨论里有说O(N²)的解法，好像是分为两个2 Sum来做。没仔细看。

下面贴个常规O(n³)解法，注意去重。 

vector<vector<int> > fourSum(vector<int> &num, int target) {

        vector<vector<int>> ans;

        if(num.size() < 4) return ans;

        int l1, l2, l3, l4;

        sort(num.begin(), num.end());

        for(l1 = 0; l1 < num.size() - 3; l1++)

        {

            if(l1 > 0 && num[l1] == num[l1 - 1]) continue; //注意去重复方法 含义是在另外三个数字固定的情况下 同一位置的数字不能重复

            for(l4 = num.size() - 1; l4 >= l1 + 3; l4--)

            {

                if(l4 < num.size() - 1 && num[l4] == num[l4 + 1]) continue;

                l2 = l1 + 1;

                l3 = l4 - 1;

                while(l2 < l3)

                {

                    if(num[l1]+num[l2]+num[l3]+num[l4] == target)

                    {

                        vector<int> partans;

                        partans.push_back(num[l1]);

                        partans.push_back(num[l2]);

                        partans.push_back(num[l3]);

                        partans.push_back(num[l4]);

                        ans.push_back(partans);

                        

                        l2++;

                        while(num[l2] == num[l2 - 1])

                        {

                            l2++;

                        }

                        l3--;

                        while(num[l3] == num[l3 + 1])

                        {

                            l3--;

                        }

                    }

                    else if(num[l1]+num[l2]+num[l3]+num[l4] < target)

                    {

                        l2++;

                    }

                    else

                    {

                        l3--;

                    }

                }

            }

        }

        return ans;

    }

 

大神看起来更统一的代码：

public class Solution {

    public List<List<Integer>> fourSum(int[] num, int target) {

        List<List<Integer>> results = new LinkedList<List<Integer>>();

        if (num == null || num.length < 4)

            return results;

        Arrays.sort(num);



        for (int s = 0; s < num.length - 3; s++) {

            if (s > 0 && num[s] == num[s - 1])  continue;





            for (int e = num.length - 1; e >= s + 3; e--) {

                if (e < num.length - 1 && num[e] == num[e + 1]) continue;



                int local = target - num[s] - num[e];

                int j = s + 1;

                int k = e - 1;

                while (j < k) {



                    if (j > s + 1 && num[j] == num[j - 1]) {

                        j++;

                        continue;

                    }

                    if (k < e - 1 && num[k] == num[k + 1]) {

                        k--;

                        continue;

                    }



                    if ((num[j] + num[k]) > local)

                        k--;

                    else if ((num[j] + num[k]) < local)

                        j++;

                    else

                        results.add(new ArrayList<Integer>(Arrays.asList(

                                num[s], num[j++], num[k--], num[e])));

                }

            }

        }

        return results;

    }

}