题目：
Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note:
- Elements in a triplet (a,b,c) must be in non-descending order. (ie, a ≤ b ≤ c)
- The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)
题解：
3 Sum是two Sum的变种，可以利用two sum的二分查找法来解决问题。
本题比two sum增加的问题有：解决duplicate问题，3个数相加返回数值而非index。
首先，对数组进行排序。
然后，从0位置开始到倒数第三个位置（num.length-3)，进行遍历，假定num[i]就是3sum中得第一个加数，然后从i+1的位置开始，进行2sum的运算。
当找到一个3sum==0的情况时，判断是否在结果hashset中出现过，没有则添加。(利用hashset的value唯一性）
因为结果不唯一，此时不能停止，继续搜索，low和high指针同时挪动。
时间复杂度是O(n2)
实现代码为：
1
public ArrayList<ArrayList<Integer>> threeSum(
int[] num) {
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(num.length<3||num == null)
4 return res;
5
6 HashSet<ArrayList<Integer>> hs = new HashSet<ArrayList<Integer>>();
7
8 Arrays.sort(num);
9
10 for( int i = 0; i <= num.length-3; i++){
11 int low = i+1;
12 int high = num.length-1;
13 while(low<high){ // since they cannot be the same one, low should not equal to high
14 int sum = num[i]+num[low]+num[high];
15 if(sum == 0){
16 ArrayList<Integer> unit = new ArrayList<Integer>();
17 unit.add(num[i]);
18 unit.add(num[low]);
19 unit.add(num[high]);
20
21 if(!hs.contains(unit)){
22 hs.add(unit);
23 res.add(unit);
24 }
25
26 low++;
27 high--;
28 } else if(sum > 0)
29 high --;
30 else
31 low ++;
32 }
33 }
34
35 return res;
36 }
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(num.length<3||num == null)
4 return res;
5
6 HashSet<ArrayList<Integer>> hs = new HashSet<ArrayList<Integer>>();
7
8 Arrays.sort(num);
9
10 for( int i = 0; i <= num.length-3; i++){
11 int low = i+1;
12 int high = num.length-1;
13 while(low<high){ // since they cannot be the same one, low should not equal to high
14 int sum = num[i]+num[low]+num[high];
15 if(sum == 0){
16 ArrayList<Integer> unit = new ArrayList<Integer>();
17 unit.add(num[i]);
18 unit.add(num[low]);
19 unit.add(num[high]);
20
21 if(!hs.contains(unit)){
22 hs.add(unit);
23 res.add(unit);
24 }
25
26 low++;
27 high--;
28 } else if(sum > 0)
29 high --;
30 else
31 low ++;
32 }
33 }
34
35 return res;
36 }
同时，解决duplicate问题，也可以通过挪动指针来解决判断，当找到一个合格结果时，将3个加数指针挪动到与当前值不同的地方，才再进行继续判断，代码如下：
1
public ArrayList<ArrayList<Integer>> threeSum(
int[] num) {
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(num.length<3||num == null)
4 return res;
5
6 Arrays.sort(num);
7
8 for( int i = 0; i <= num.length-3; i++){
9 if(i==0||num[i]!=num[i-1]){ // remove dupicate
10 int low = i+1;
11 int high = num.length-1;
12 while(low<high){
13 int sum = num[i]+num[low]+num[high];
14 if(sum == 0){
15 ArrayList<Integer> unit = new ArrayList<Integer>();
16 unit.add(num[i]);
17 unit.add(num[low]);
18 unit.add(num[high]);
19
20 res.add(unit);
21
22 low++;
23 high--;
24
25 while(low<high&&num[low]==num[low-1]) // remove dupicate
26 low++;
27 while(low<high&&num[high]==num[high+1]) // remove dupicate
28 high--;
29
30 } else if(sum > 0)
31 high --;
32 else
33 low ++;
34 }
35 }
36 }
37 return res;
38 }
2 ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
3 if(num.length<3||num == null)
4 return res;
5
6 Arrays.sort(num);
7
8 for( int i = 0; i <= num.length-3; i++){
9 if(i==0||num[i]!=num[i-1]){ // remove dupicate
10 int low = i+1;
11 int high = num.length-1;
12 while(low<high){
13 int sum = num[i]+num[low]+num[high];
14 if(sum == 0){
15 ArrayList<Integer> unit = new ArrayList<Integer>();
16 unit.add(num[i]);
17 unit.add(num[low]);
18 unit.add(num[high]);
19
20 res.add(unit);
21
22 low++;
23 high--;
24
25 while(low<high&&num[low]==num[low-1]) // remove dupicate
26 low++;
27 while(low<high&&num[high]==num[high+1]) // remove dupicate
28 high--;
29
30 } else if(sum > 0)
31 high --;
32 else
33 low ++;
34 }
35 }
36 }
37 return res;
38 }