3Sum

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)

思想： 把3sum转换为2Sum，并且查找时避免重复，将重复项去除。

1. public ArrayList<ArrayList<Integer>> threeSum(int[] num) {
2. ArrayList<ArrayList<Integer> > res = new ArrayList<ArrayList<Integer> >();
3. if(num == null || num.length < 3) return res;
4. int lg = num.length;
5. Arrays.sort(num);
6. for(int i=0;i<=lg-3;i++) {
7. if(i>0 && num[i] == num[i-1]) continue;
8. ArrayList<Integer> cur = new ArrayList<Integer>();
9. int target = 0-num[i];
10. int low = i+1;
11. int high = lg-1;
12. while(low<high) {
13. if(num[low]+num[high]==target) {
14. cur.add(num[i]);
15. cur.add(num[low]);
16. cur.add(num[high]);
17. res.add(new ArrayList<Integer>(cur));
18. low++;
19. while(low<high && num[low]==num[low-1]) low++;
20. high--;
21. while(high>low && num[high]==num[high+1]) high--;
22. cur.clear();
23. } else if(num[low]+num[high]>target) {
24. high--;
25. while(high>low && num[high] == num[high+1]) high--;
26. } else {
27. low++;
28. while(low<high && num[low]==num[low-1]) low++;
29. }
30. }
31. }
32. return res;
33. }