[LeetCode] 18. 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: 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] ]
Thought process:
Add one loop to "15. 3Sum".
Solution:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 | public class Solution { public List<List<Integer>> fourSum(int[] nums, int target) { Arrays.sort(nums); List<List<Integer>> quadruplets = new ArrayList<>(); for (int i = 0; i < nums.length - 3; i++) { while (i > 0 && i < nums.length - 3 && nums[i] == nums[i - 1]) { i++; } for (int j = i + 1; j < nums.length - 2; j++) { while (j > i + 1 && j < nums.length - 2 && nums[j] == nums[j - 1]) { j++; } if (j >= nums.length) { break; } int start = j + 1; int end = nums.length - 1; while (start < end) { int fourSum = nums[i] + nums[j] + nums[start] + nums[end]; if (fourSum < target) { start++; } else if (fourSum > target) { end--; } else { List<Integer> quadruplet = new ArrayList<>(); quadruplet.addAll(Arrays.asList(nums[i], nums[j], nums[start], nums[end])); quadruplets.add(quadruplet); start++; while (start < nums.length && nums[start] == nums[start - 1]) { start++; } end--; while (end >= 0 && nums[end] == nums[end + 1]) { end--; } } } } } return quadruplets; } } |
Time complexity:
Sorting takes O(nlogn). Loop takes O(n^3). The overall time complexity is O(n^3).
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