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] 3Sum的扩展,在3Sum外面再加一层循环,注意去重处理。
public class Solution { public List<List<Integer>> fourSum(int[] nums, int target) { List<List<Integer>> result = new ArrayList<List<Integer>>(); if (nums == null || nums.length < 4) return result; Arrays.sort(nums); int N = nums.length; for (int i = 0; i < N - 3; i++) { if (i > 0 && nums[i] == nums[i - 1]) continue; for (int j = i + 1; j < N - 2; j++) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int p = j + 1, q = N - 1; while (p < q) { int sum4 = nums[i] + nums[j] + nums[p] + nums[q]; if (sum4 == target) { List<Integer> item = new ArrayList<Integer>(); item.add(nums[i]); item.add(nums[j]); item.add(nums[p]); item.add(nums[q]); result.add(item); while (++p < q && nums[p] == nums[p - 1]); while (p < --q && nums[q] == nums[q + 1]); } else if (sum4 > target) { while (p < --q && nums[q] == nums[q + 1]); } else { while (++p < q && nums[p] == nums[p - 1]); } } } } return result; } }