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]
]
1 public List<List<Integer>> fourSum(int[] nums, int target) { 2 List<List<Integer>> res = new ArrayList<>(); 3 Arrays.sort(nums); 4 for (int i = 0; i < nums.length - 3; i++) { 5 if (i != 0 && nums[i] == nums[i - 1]) 6 continue; 7 for (int j = i + 1; j < nums.length - 2; j++) { 8 if (j > i + 1 && nums[j] == nums[j - 1]) 9 continue; 10 int k = j + 1; 11 int l = nums.length - 1; 12 while (k < l) { 13 int sum = nums[i] + nums[j] + nums[k] + nums[l]; 14 if (sum == target) { 15 List<Integer> list = new ArrayList<>(); 16 list.add(nums[i]); 17 list.add(nums[j]); 18 list.add(nums[k]); 19 list.add(nums[l]); 20 res.add(list); 21 k++; 22 l--; 23 // 去重复 24 while (k < l && nums[k] == nums[k - 1]) { 25 k++; 26 } 27 while (k < l && nums[l] == nums[l + 1]) { 28 l--; 29 } 30 } else if (sum < target) { 31 k++; 32 } else { 33 l--; 34 } 35 } 36 } 37 } 38 return res; 39 }
1 public List<List<Integer>> fourSum3(int[] num, int target) { 2 ArrayList<List<Integer>> ans = new ArrayList<>(); 3 if (num.length < 4) 4 return ans; 5 Arrays.sort(num); 6 for (int i = 0; i < num.length - 3; i++) { 7 if (num[i] + num[i + 1] + num[i + 2] + num[i + 3] > target) 8 break; // first candidate too large, search finished 9 if (num[i] + num[num.length - 1] + num[num.length - 2] + num[num.length - 3] < target) 10 continue; // first candidate too small 11 if (i > 0 && num[i] == num[i - 1]) 12 continue; // prevents duplicate result in ans list 13 for (int j = i + 1; j < num.length - 2; j++) { 14 if (num[i] + num[j] + num[j + 1] + num[j + 2] > target) 15 break; // second candidate too large 16 if (num[i] + num[j] + num[num.length - 1] + num[num.length - 2] < target) 17 continue; // second candidate too small 18 if (j > i + 1 && num[j] == num[j - 1]) 19 continue; // prevents duplicate results in ans list 20 int low = j + 1, high = num.length - 1; 21 while (low < high) { 22 int sum = num[i] + num[j] + num[low] + num[high]; 23 if (sum == target) { 24 ans.add(Arrays.asList(num[i], num[j], num[low], num[high])); 25 while (low < high && num[low] == num[low + 1]) 26 low++; // skipping over duplicate on low 27 while (low < high && num[high] == num[high - 1]) 28 high--; // skipping over duplicate on high 29 low++; 30 high--; 31 } 32 // move window 33 else if (sum < target) 34 low++; 35 else 36 high--; 37 } 38 } 39 } 40 return ans; 41 }