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] ]
题目含义:从数组中找4个数字,使得加和等于目标值target,找出所有的组合
1 public List<List<Integer>> threeSum(int[] nums, int target, int firstNumber) { 2 List<List<Integer>> res = new LinkedList<>(); 3 for (int i = 0; i < nums.length; i++) { 4 if (i > 0 && nums[i] == nums[i - 1]) continue; 5 int low = i + 1, high = nums.length - 1; 6 while (low < high) { 7 if (nums[i] + nums[low] + nums[high] == target) { 8 res.add(Arrays.asList(firstNumber, nums[i], nums[low], nums[high])); 9 while (low < high && nums[low] == nums[low + 1]) low++; 10 while (low < high && nums[high] == nums[high - 1]) high--; 11 low++; 12 high--; 13 } else if (nums[i] + nums[low] + nums[high] > target) high--; 14 else low++; 15 } 16 } 17 return res; 18 } 20 21 public List<List<Integer>> fourSum(int[] nums, int target) { 22 Arrays.sort(nums); 23 List<List<Integer>> res = new LinkedList<>(); 24 if (nums.length == 0) return res; 25 for (int i = 0; i < nums.length; i++) { 26 if (i > 0 && nums[i] == nums[i - 1]) continue; 27 int[] rightNums = Arrays.copyOfRange(nums, i+1, nums.length); 28 List<List<Integer>> threeSum = threeSum(rightNums, target - nums[i], nums[i]); 29 res.addAll(threeSum); 30 } 31 return res; 32 }
类似题目:15. 3Sum
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