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]]这道题跟3SUM类似，多加一层循环即可。先对数组进行排序，前两个数用循环，后两个数用two point从两边向中间逼近，同时注意跳过重复的情况。代码如下：
public class Solution {    public List<List<Integer>> fourSum(int[] nums, int target) {        Arrays.sort(nums);        List<List<Integer>> res = new ArrayList<List<Integer>>();        for (int i = 0; i < nums.length - 3; i ++) {            if (i == 0 || nums[i] != nums[i - 1]) {                for (int j = i + 1; j < nums.length - 2; j ++) {                    if (j == i + 1 || nums[j] != nums[j - 1]) {                        int left = j + 1, right = nums.length - 1, sum = target - nums[i] - nums[j];                        while (left < right) {                            if (nums[left] + nums[right] == sum) {                                List<Integer> tempList = new ArrayList<Integer>();                                tempList.add(nums[i]);                                tempList.add(nums[j]);                                tempList.add(nums[left]);                                tempList.add(nums[right]);                                res.add(tempList);                                left ++;                                right --;                                while (left < right && nums[left] == nums[left - 1]) left ++;                                while (left < right && nums[right] == nums[right + 1]) right --;                            } else if (nums[left] + nums[right] < sum) {                                left ++;                            } else {                                right --;                            }                        }                    }                }            }        }        return res;    }}另外，beat 100%的方法踢掉了很多不需要的部分，可以借鉴他的限制条件，代码如下：
public List<List<Integer>> fourSum(int[] nums, int target) {		ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();		int len = nums.length;		if (nums == null || len < 4)			return res;		Arrays.sort(nums);		int max = nums[len - 1];		if (4 * nums[0] > target || 4 * max < target)			return res;		int i, z;		for (i = 0; i < len; i++) {			z = nums[i];			if (i > 0 && z == nums[i - 1])// avoid duplicate				continue;			if (z + 3 * max < target) // z is too small				continue;			if (4 * z > target) // z is too large				break;			if (4 * z == target) { // z is the boundary				if (i + 3 < len && nums[i + 3] == z)					res.add(Arrays.asList(z, z, z, z));				break;			}			threeSumForFourSum(nums, target - z, i + 1, len - 1, res, z);		}		return res;	}	/*	 * Find all possible distinguished three numbers adding up to the target	 * in sorted array nums[] between indices low and high. If there are,	 * add all of them into the ArrayList fourSumList, using	 * fourSumList.add(Arrays.asList(z1, the three numbers))	 */	public void threeSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,			int z1) {		if (low + 1 >= high)			return;		int max = nums[high];		if (3 * nums[low] > target || 3 * max < target)			return;		int i, z;		for (i = low; i < high - 1; i++) {			z = nums[i];			if (i > low && z == nums[i - 1]) // avoid duplicate				continue;			if (z + 2 * max < target) // z is too small				continue;			if (3 * z > target) // z is too large				break;			if (3 * z == target) { // z is the boundary				if (i + 1 < high && nums[i + 2] == z)					fourSumList.add(Arrays.asList(z1, z, z, z));				break;			}			twoSumForFourSum(nums, target - z, i + 1, high, fourSumList, z1, z);		}	}	/*	 * Find all possible distinguished two numbers adding up to the target	 * in sorted array nums[] between indices low and high. If there are,	 * add all of them into the ArrayList fourSumList, using	 * fourSumList.add(Arrays.asList(z1, z2, the two numbers))	 */	public void twoSumForFourSum(int[] nums, int target, int low, int high, ArrayList<List<Integer>> fourSumList,			int z1, int z2) {		if (low >= high)			return;		if (2 * nums[low] > target || 2 * nums[high] < target)			return;		int i = low, j = high, sum, x;		while (i < j) {			sum = nums[i] + nums[j];			if (sum == target) {				fourSumList.add(Arrays.asList(z1, z2, nums[i], nums[j]));				x = nums[i];				while (++i < j && x == nums[i]) // avoid duplicate					;				x = nums[j];				while (i < --j && x == nums[j]) // avoid duplicate					;			}			if (sum < target)				i++;			if (sum > target)				j--;		}		return;	}}