问题描述:
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)
解题思路:
与之前的3sum相类似,只不过在外边多了一套循环,时间复杂度为O(n^3)
代码如下:
public class Solution { List<List<Integer>> ans = new ArrayList<List<Integer>>(); public List<List<Integer>> fourSum(int[] nums, int target) { int length = nums.length; int tmp; if (nums == null || length < 4) return ans; Arrays.sort(nums); for (int i = 0; i < length - 3; i++) { if (i > 0 && nums[i] == nums[i - 1]) continue; for (int j = i + 1; j < length - 2; j++) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int begin = j + 1; int end = length - 1; while (begin < end) { tmp = nums[begin] + nums[end] + nums[i] + nums[j]; if (tmp == target) { List<Integer> list = new ArrayList<Integer>(); list.add(nums[i]); list.add(nums[j]); list.add(nums[begin]); list.add(nums[end]); ans.add(list); while (begin < end && nums[begin + 1] == nums[begin]) begin++; begin++; while (begin < end && nums[end - 1] == nums[end]) end--; end--; } else if (tmp > target) end--; else begin++; } } } return ans; } }