*4sum

题目：

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.

    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)
 题解： 4 sum跟3 sum是一样的思路，只不过需要多考虑一个加数，这样时间复杂度变为O(n3)。
使用HashSet来解决重复问题的代码如下：

 1 public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
 2     HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
 3     ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 4     Arrays.sort(num);
 5     for (int i = 0; i <= num.length-4; i++) {
 6         for (int j = i + 1; j <= num.length-3; j++) {
 7             int low = j + 1;
 8             int high = num.length - 1;
 9  
10             while (low < high) {
11                 int sum = num[i] + num[j] + num[low] + num[high];
12  
13                 if (sum > target) {
14                     high--;
15                 } else if (sum < target) {
16                     low++;
17                 } else if (sum == target) {
18                     ArrayList<Integer> temp = new ArrayList<Integer>();
19                     temp.add(num[i]);
20                     temp.add(num[j]);
21                     temp.add(num[low]);
22                     temp.add(num[high]);
23  
24                     if (!hashSet.contains(temp)) {
25                         hashSet.add(temp);
26                         result.add(temp);
27                     }
28  
29                     low++;
30                     high--;
31                 }
32             }
33         }
34     }
35  
36     return result;
37 }

使用挪动指针的方法来解决重复的代码如下：

 1 public ArrayList<ArrayList<Integer>> fourSum(int[] num, int target) {
 2     HashSet<ArrayList<Integer>> hashSet = new HashSet<ArrayList<Integer>>();
 3     ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
 4     Arrays.sort(num);
 5     for (int i = 0; i <= num.length-4; i++) {
 6         if(i==0||num[i]!=num[i-1]){
 7             for (int j = i + 1; j <= num.length-3; j++) {
 8                 if(j==i+1||num[j]!=num[j-1]){
 9                     int low = j + 1;
10                     int high = num.length - 1;
11          
12                     while (low < high) {
13                         int sum = num[i] + num[j] + num[low] + num[high];
14          
15                         if (sum > target) {
16                             high--;
17                         } else if (sum < target) {
18                             low++;
19                         } else if (sum == target) {
20                             ArrayList<Integer> temp = new ArrayList<Integer>();
21                             temp.add(num[i]);
22                             temp.add(num[j]);
23                             temp.add(num[low]);
24                             temp.add(num[high]);
25          
26                             if (!hashSet.contains(temp)) {
27                                 hashSet.add(temp);
28                                 result.add(temp);
29                             }
30          
31                             low++;
32                             high--;
33                             
34                             while(low<high&&num[low]==num[low-1])//remove dupicate
35                                 low++;
36                             while(low<high&&num[high]==num[high+1])//remove dupicate
37                                 high--;
38                         }
39                     }
40                 }
41             }
42         }
43     }
44  
45     return result;
46 }