LeetCode - 18. 4Sum

解题代码:

classSolution {

public:

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

        vector<vector<int>> res;

        if(nums.size()<4)

            return res;

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

        for(int i=0;i<nums.size()-3;i++){

            int t=target-nums[i];

            for(intj=i+1;j<nums.size();j++){

                int a=j+1,b=nums.size()-1;

                while(a<b){

                    intsum=nums[j]+nums[a]+nums;

                    if(sum==t){

                        vector<int> d(4,0);

                        d[0]=nums[i];

                        d[1]=nums[j];

                        d[2]=nums[a];

                        d[3]=nums[b];

                        res.push_back(d);

                       while(a<b&&nums[b]==d[3])

                            b--;

                       while(a<b&&nums[a]==d[2])

                            a++;

                    }

                    else if(sum<t)

                        a++;

                    else

                        b--;

                }

               while(j<nums.size()-1&&nums[j+1]==nums[j])

                    j++;

            }

           while(i<nums.size()-1&&nums[i+1]==nums[i])

                i++;

        }

        return res;

    }

};

 

[b]解题思路:

本题与前面15题3 SUM可以说几乎没有太多区别,尽管多了一个数,但思路完全一样,对于数组中每个数num[i],相当于求3 sum 的结果为target- num[i]。因此可以用几乎一样的方法进行求解。