3 Sum 解答

Question

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)

Solution

First, we sort the array and then implement 2 pointers to get 2 Sum. Time complexity is O(n²)

Notice here, we need to consider duplicates in result.

 public class Solution {
     public List<List<Integer>> threeSum(int[] nums) {
         if (nums == null || nums.length < 1)
             return null;
         Arrays.sort(nums);
         int length = nums.length;
         List<List<Integer>> result = new ArrayList<List<Integer>>();
         for (int i = 0; i <= length - 3; i++) {
             // Remove duplicates
             if (i > 0 && nums[i] == nums[i - 1])
                 continue;
             int target = 0 - nums[i];
             int l = i + 1, r = length - 1;
             while (l < r) {
                 if (nums[l] + nums[r] == target) {
                     result.add(Arrays.asList(nums[i], nums[l], nums[r]));
                     while (l < r && nums[l] == nums[l+1]) l++;
                     while (l < r && nums[r] == nums[r-1]) r--;
                     l++;
                     r--;
                 } else if (nums[l] + nums[r] < target) {
                     l++;
                 } else {
                     r--;
                 }
             }
         }
         return result;
     }
 }