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

  public List<List<Integer>> threeSum(int[] nums, int target, int firstNumber) {
        List<List<Integer>> res = new LinkedList<>();
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int low = i + 1, high = nums.length - 1;
            while (low < high) {
                if (nums[i] + nums[low] + nums[high] == target) {
                    res.add(Arrays.asList(firstNumber, nums[i], nums[low], nums[high]));
                    while (low < high && nums[low] == nums[low + 1]) low++;
                    while (low < high && nums[high] == nums[high - 1]) high--;
                    low++;
                    high--;
                } else if (nums[i] + nums[low] + nums[high] > target) high--;
                else low++;
            }
        }
        return res;
    }

    public List<List<Integer>> fourSum(int[] nums, int target) {
        Arrays.sort(nums);
        List<List<Integer>> res = new LinkedList<>();
        if (nums.length == 0) return res;
        for (int i = 0; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1]) continue;
            int[] rightNums = Arrays.copyOfRange(nums, i+1, nums.length);
            List<List<Integer>> threeSum = threeSum(rightNums, target - nums[i], nums[i]);
            res.addAll(threeSum);
        }
        return res;
    }

 类似题目：15. 3Sum