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)

 Idea:  o(n^2) solution exists. First sort the array, and then from left to right, for each num[i], search the pair that sums up to -num[i] using Two Sum algorithm. 

 1 public class Solution {

 2     public ArrayList<ArrayList<Integer>> threeSum(int[] num) {

 3         // Note: The Solution object is instantiated only once and is reused by each test case.

 4         Arrays.sort(num);

 5         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

 6         if(num == null || num.length < 3) return result;

 7         int len = num.length;

 8         for(int i = 0; i < num.length-2; i ++){

 9             if(i==0 || num[i]>num[i-1]){

10                 int j = i + 1;

11                 int h = len - 1;

12                 while(j < h){

13                     int sum = 0 - num[j] - num[h];

14                     if(sum == num[i]){

15                         ArrayList<Integer> row = new ArrayList<Integer>();

16                         row.add(num[i]);

17                         row.add(num[j]);

18                         row.add(num[h]);

19                         result.add(row);

20                         h--;

21                         j++;

22                         while(h>j && num[h]==num[h+1]) h--; 

23 

24                         while(j<h && num[j]==num[j-1]) j++;

25                     }else if(sum < num[i]){

26                             h --;

27                     }else if(sum > num[i]){

28                             j ++;

29                     }

30                 }

31             }

32         }

33         return result;

34     }

35 }

 

 第二遍：

 1 public class Solution {

 2     public ArrayList<ArrayList<Integer>> threeSum(int[] num) {

 3         // Note: The Solution object is instantiated only once and is reused by each test case.

 4         ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

 5         if(num == null || num.length < 3) return result;

 6         Arrays.sort(num);

 7         for(int i = 0; i < num.length - 2; i ++){

 8             if(i == 0 || num[i] > num[i-1]){

 9                 int target = - num[i];

10                 int j = i + 1;

11                 int h = num.length - 1;

12                 while(j < h){

13                     if(num[j] + num[h] == target){

14                         ArrayList<Integer> row = new ArrayList<Integer>();

15                         row.add(num[i]);

16                         row.add(num[j]);

17                         row.add(num[h]);

18                         result.add(row);

19                         int jnum = num[j];

20                         int hnum = num[h];

21                         while(num[j] == jnum && j < h) j ++;

22                         while(num[h] == hnum && h > j) h --;

23                     }else if(num[j] + num[h] > target){

24                         h --;

25                     }else{

26                         j ++;

27                     }

28                 }

29             }

30         }

31         return result;

32     }

33 }