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)

給一個含有n個整數的數組S， 找出所有不重復的4個數和等於target的所有解。

解法1：Two pointers, 跟3Sum差不多，只是擴展到4個數字的和。還是按照3Sum的解法，先排序，只是在外面套一層循環，相當於求n次3Sum，用set()去重。3Sum的時間復雜度是O(n^2)，所以總時間復雜度O(n^3)。但這個時間復雜度高，用時太長。

解法2：Hash map, 以空間換時間，首先建立一個Hash map，key值為數組中每兩個元素的和，對應的value為這兩個元素的下標組成的元組num[p]+num[q]] : (p,q) ，pairs元組不一定是唯一的。如對於num=[1,2,3,2]來說，dict={3:[(0,1),(0,3)], 4:[(0,2),(1,3)], 5:[(1,2),(2,3)]}。這樣就可以檢查target-key這個值在不在dict的key值中，如果target-key在dict中並且下標符合要求，那麽就找到了這樣的一組解。用set()去重。

Java: 2 pointers

public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
    Arrays.sort(num);  
    HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
    ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
  
    for (int i = 0; i < num.length; i++) {
        for (int j = i + 1; j < num.length; j++) {
            int k = j + 1;
            int l = num.length - 1;
  
            while (k < l) {
                int sum = num[i] + num[j] + num[k] + num[l];
  
                if (sum > target) {
                    l--;
                } else if (sum < target) {
                    k++;
                } else if (sum == target) {
                    ArrayList<Integer> temp = new ArrayList<Integer>();
                    temp.add(num[i]);
                    temp.add(num[j]);
                    temp.add(num[k]);
                    temp.add(num[l]);
  
                    if (!hashSet.contains(temp)) {
                        hashSet.add(temp);
                        result.add(temp);
                    }
  
                    k++;
                    l--;
                }
            }
        }
    }
  
    return result;
}

Python: Two pointers, Time: O(n^3), Space: O(1)

# Two pointer solution. (1356ms)
class Solution(object):
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        nums.sort()
        res = []
        for i in xrange(len(nums) - 3):
            if i and nums[i] == nums[i - 1]:
                continue
            for j in xrange(i + 1, len(nums) - 2):
                if j != i + 1 and nums[j] == nums[j - 1]:
                    continue
                sum = target - nums[i] - nums[j]
                left, right = j + 1, len(nums) - 1
                while left < right:
                    if nums[left] + nums[right] == sum:
                        res.append([nums[i], nums[j], nums[left], nums[right]])
                        right -= 1
                        left += 1
                        while left < right and nums[left] == nums[left - 1]:
                            left += 1
                        while left < right and nums[right] == nums[right + 1]:
                            right -= 1
                    elif nums[left] + nums[right] > sum:
                        right -= 1
                    else:
                        left += 1
        return res

Python: HashMap, Time: O(n^2 * p), Space: O(n^2 * p)

# Hash solution. (224ms)
class Solution2(object):
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        nums, result, lookup = sorted(nums), [], collections.defaultdict(list)
        for i in xrange(0, len(nums) - 1):
            for j in xrange(i + 1, len(nums)): 
                is_duplicated = False
                for [x, y] in lookup[nums[i] + nums[j]]:
                    if nums[x] == nums[i]:
                        is_duplicated = True
                        break
                if not is_duplicated:
                    lookup[nums[i] + nums[j]].append([i, j])
        ans = {}
        for c in xrange(2, len(nums)):
            for d in xrange(c+1, len(nums)):
                if target - nums[c] - nums[d] in lookup:
                    for [a, b] in lookup[target - nums[c] - nums[d]]:
                        if b < c:
                            quad = [nums[a], nums[b], nums[c], nums[d]]
                            quad_hash = " ".join(str(quad))
                            if quad_hash not in ans:
                                ans[quad_hash] = True
                                result.append(quad)
        return result

C++: 2 pointers

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &nums, int target) {
        set<vector<int> > res;
        sort(nums.begin(), nums.end());
        for (int i = 0; i < int(nums.size() - 3); ++i) {
            for (int j = i + 1; j < int(nums.size() - 2); ++j) {
                int left = j + 1, right = nums.size() - 1;
                while (left < right) {
                    int sum = nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        vector<int> out;
                        out.push_back(nums[i]);
                        out.push_back(nums[j]);
                        out.push_back(nums[left]);
                        out.push_back(nums[right]);
                        res.insert(out);
                        ++left; --right;
                    } else if (sum < target) ++left;
                    else --right;
                }
            }
        }
        return vector<vector<int> > (res.begin(), res.end());
    }
};

[LeetCode] 18. 4Sum 四數之和