[LeetCode] 18. 4Sum 四数之和

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

The solution set must not contain duplicate quadruplets.

Example:

Given array nums = [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]
]

给一个含有n个整数的数组S， 找出所有不重复的4个数和等于target的所有解。

解法1：Two pointers, 跟3Sum差不多，只是扩展到4个数字的和。还是按照3Sum的解法，先排序，只是在外面套一层循环，相当于求n次3Sum，用set()去重。3Sum的时间复杂度是O(n^2)，所以总时间复杂度O(n^3)。但这个时间复杂度高，用时太长。

解法2：Hash map, 以空间换时间，首先建立一个Hash map，key值为数组中每两个元素的和，对应的value为这两个元素的下标组成的元组num[p]+num[q]] : (p,q) ，pairs元组不一定是唯一的。如对于num=[1,2,3,2]来说，dict={3:[(0,1),(0,3)], 4:[(0,2),(1,3)], 5:[(1,2),(2,3)]}。这样就可以检查target-key这个值在不在dict的key值中，如果target-key在dict中并且下标符合要求，那么就找到了这样的一组解。用set()去重。

Java: 2 pointers

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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;
}

Java:

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public class Solution {
        int len = 0;
        public List<List<Integer>> fourSum(int[] nums, int target) {
            len = nums.length;
            Arrays.sort(nums);
            return kSum(nums, target, 4, 0);
        }
       private ArrayList<List<Integer>> kSum(int[] nums, int target, int k, int index) {
            ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
            if(index >= len) {
                return res;
            }
            if(k == 2) {
                int i = index, j = len - 1;
                while(i < j) {
                    //find a pair
                    if(target - nums[i] == nums[j]) {
                        List<Integer> temp = new ArrayList<>();
                        temp.add(nums[i]);
                        temp.add(target-nums[i]);
                        res.add(temp);
                        //skip duplication
                        while(i<j && nums[i]==nums[i+1]) i++;
                        while(i<j && nums[j-1]==nums[j]) j--;
                        i++;
                        j--;
                    //move left bound
                    } else if (target - nums[i] > nums[j]) {
                        i++;
                    //move right bound
                    } else {
                        j--;
                    }
                }
            } else{
                for (int i = index; i < len - k + 1; i++) {
                    //use current number to reduce ksum into k-1sum
                    ArrayList<List<Integer>> temp = kSum(nums, target - nums[i], k-1, i+1);
                    if(temp != null){
                        //add previous results
                        for (List<Integer> t : temp) {
                            t.add(0, nums[i]);
                        }
                        res.addAll(temp);
                    }
                    while (i < len-1 && nums[i] == nums[i+1]) {
                        //skip duplicated numbers
                        i++;
                    }
                }
            }
            return res;
        }
    }　　

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

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# 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)

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# 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

Python:

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def fourSum(self, nums, target):
    def findNsum(nums, target, N, result, results):
        if len(nums) < N or N < 2 or target < nums[0]*N or target > nums[-1]*N:  # early termination
            return
        if N == 2: # two pointers solve sorted 2-sum problem
            l,r = 0,len(nums)-1
            while l < r:
                s = nums[l] + nums[r]
                if s == target:
                    results.append(result + [nums[l], nums[r]])
                    l += 1
                    while l < r and nums[l] == nums[l-1]:
                        l += 1
                elif s < target:
                    l += 1
                else:
                    r -= 1
        else: # recursively reduce N
            for i in range(len(nums)-N+1):
                if i == 0 or (i > 0 and nums[i-1] != nums[i]):
                    findNsum(nums[i+1:], target-nums[i], N-1, result+[nums[i]], results)
 
    results = []
    findNsum(sorted(nums), target, 4, [], results)
    return results　　

C++: 2 pointers

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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());
    }
};

 

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[LeetCode] 454. 4Sum II 四数之和II

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