4 Sum

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.

Example

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：看到提示是使用Hashmap来做，没想到怎么处理。自己实现了全部迭代的解法，全部迭代因为不会出现重复计算，所以计算量也还是可以控制。参考网上的解法，实现双指针夹逼解法。

解法一：

public class Solution {    /**     * @param numbers : Give an array numbersbers of n integer     * @param target : you need to find four elements that's sum of target     * @return : Find all unique quadruplets in the array which gives the sum of     *           zero.     */    public ArrayList<ArrayList<Integer>> fourSum(int[] numbers, int target) {Arrays.sort(numbers);ArrayList<ArrayList<Integer>> res = find(numbers, target, 0, 4);return res == null ? new ArrayList<ArrayList<Integer>>() : res;}public ArrayList<ArrayList<Integer>> find(int[] numbers, int target, int i, int n) {int N = numbers.length;if (n == 1) {while (i < N && numbers[i] != target)i++;if (i < N) {ArrayList<Integer> l = new ArrayList<Integer>();l.add(numbers[i]);ArrayList<ArrayList<Integer>> list = new ArrayList<ArrayList<Integer>>();list.add(l);return list;} elsereturn null;} else {int t = i;ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();while (t + n <= N) {if (t > i && numbers[t] == numbers[t - 1]) {t++;continue;}target = target - numbers[t];ArrayList<ArrayList<Integer>> list = find(numbers, target, t + 1, n - 1);if (list != null) {for (ArrayList l : list) {l.add(0, numbers[t]);res.add(l);}}target = target + numbers[t];t++;}if (res.size() > 0)return res;elsereturn null;}}    }

解法二：

public class Solution {    /**     * @param numbers : Give an array numbersbers of n integer     * @param target : you need to find four elements that's sum of target     * @return : Find all unique quadruplets in the array which gives the sum of     *           zero.     */    public ArrayList<ArrayList<Integer>> fourSum(int[] numbers, int target) {Arrays.sort(numbers);ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();if (numbers == null || numbers.length < 4)return res;int N = numbers.length;for (int i = 0; i < N - 3; i++) {for (int j = i + 1; j < N - 2; j++) {int l = j + 1;int h = N - 1;while (l < h) {int sum = numbers[i] + numbers[j] + numbers[l] + numbers[h];if (sum < target)l++;else if (sum > target)h--;else {boolean find = false;for (ArrayList<Integer> ll : res) {if (ll.get(0) == numbers[i] && ll.get(1) == numbers[j] && ll.get(2) == numbers[l]&& ll.get(3) == numbers[h])find = true;}if (!find) {ArrayList<Integer> ll = new ArrayList<Integer>();ll.add(numbers[i]);ll.add(numbers[j]);ll.add(numbers[l]);ll.add(numbers[h]);res.add(ll);}l++;h--;}}}}return res;}}