4Sum

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)

分析：和3Sum的做法类似，将数组排序后，固定第一、二个数，然后从两边夹逼第三、四个数。注意检查重复数字出现的情况。

Solution1: 运行时间129ms。

class Solution {
public:
    vector<vector<int> > fourSum(vector<int>& nums, int target) {
        vector<vector<int> >result;
        if(nums.size() < 4) return result;
        sort(nums.begin(), nums.end());
        
        for(int a = 0; a < nums.size() - 3; a++){
            if(a != 0 && nums[a] == nums[a-1]) continue;
            for(int b = a + 1; b < nums.size() - 2; b++){
                if(b != a + 1 && nums[b] == nums[b-1]) continue;
                int c = b + 1, d = nums.size() - 1;
                
                while(c < d){
                    if(nums[a] + nums[b] + nums[c] + nums[d] > target) d--;
                    else if(nums[a] + nums[b] + nums[c] + nums[d] < target) c++;
                    else{
                        if(c != b + 1 && nums[c] == nums[c-1]) c++;
                        else if(d != nums.size() - 1 && nums[d] == nums[d+1]) d--;
                        else{
                            result.push_back({nums[a], nums[b], nums[c], nums[d]});
                            c++;
                            d--;
                        }
                    }
                }
            }
        }
        sort(result.begin(), result.end());
        return result;
    }
};

Solution2:用map进行存储与判断，耗时更多。160ms。不推荐。

class Solution {
public:
    vector<vector<int> > fourSum(vector<int>& nums, int target) {
        vector<vector<int> >result;
        if(nums.size() < 4) return result;
        sort(nums.begin(), nums.end());
        
        for(int a = 0; a < nums.size() - 3; a++){
            if(a != 0 && nums[a] == nums[a-1]) continue;
            for(int b = a + 1; b < nums.size() - 2; b++){
                if(b != a + 1 && nums[b] == nums[b-1]) continue;
                int c = b + 1, d = nums.size() - 1;
                map<int, int> hmap;
                
                while(c < d){
                    if(nums[a] + nums[b] + nums[c] + nums[d] > target) d--;
                    else if(nums[a] + nums[b] + nums[c] + nums[d] < target) c++;
                    else{
                        if(hmap.find(nums[c]) == hmap.end()){
                            hmap[nums[c]] = nums[d];
                            result.push_back({nums[a], nums[b], nums[c], nums[d]});
                        }
                        c++;
                        d--;
                    }
                }
            }
        }
        sort(result.begin(), result.end());
        return result;
    }
};