leetcode18~4Sum

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.
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]    ]

比求3Sum多一层嵌套循环，还有一种解法是使用hashmap，先缓存两个数的和，然后再求解。
下面给出了两种解法，第二种解法参考leetcode讨论区的代码，过滤条件比较多，最后运行时间真的很短，可以借鉴一下~

public class FourSum {    //时间复杂度O(n^3)    public List<List<Integer>> fourSum2(int[] nums, int target) {        List<List<Integer>> res = new ArrayList<>();        if(nums==null || nums.length<4) return res;        Arrays.sort(nums);        for(int i=0;i<nums.length-3;i++) {            if(i>0 && nums[i]==nums[i-1]) continue;            for(int j=i+1;j<nums.length-2;j++) {                if(j>i+1 && nums[j]==nums[j-1]) continue;                int top = j+1,tail= nums.length-1;                while(top<tail) {                    if(nums[i]+nums[j]+nums[top]+nums[tail]==target) {                        List<Integer> list = new ArrayList<Integer>();                        list.add(nums[i]);                        list.add(nums[j]);                        list.add(nums[top]);                        list.add(nums[tail]);                        res.add(list);                        while(top<tail && nums[top]==nums[top+1]) top++;                        while(top<tail && nums[tail]==nums[tail-1]) tail--;                        top++;                        tail--;                    } else if(nums[i]+nums[j]+nums[top]+nums[tail]<target) {                        top++;                    } else {                        tail--;                    }                }            }        }        return res;    }    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;    }}