Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
Notice that the solution set must not contain duplicate triplets.
Example 1: Input: nums = [-1,0,1,2,-1,-4] Output: [[-1,-1,2],[-1,0,1]] Explanation: nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0. nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0. nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0. The distinct triplets are [-1,0,1] and [-1,-1,2]. Notice that the order of the output and the order of the triplets does not matter.
Example 2: Input: nums = [0,1,1] Output: [] Explanation: The only possible triplet does not sum up to 0.
Example 3: Input: nums = [0,0,0] Output: [[0,0,0]] Explanation: The only possible triplet sums up to 0.
Constraints:
3 <= nums.length <= 3000-105 <= nums[i] <= 105
Solution
Java
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> list = new ArrayList<List<Integer>>();
Arrays.sort(nums);
for(int i = 0 ; i < nums.length - 2; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int l = i + 1;
int r = nums.length - 1;
int target = 0 - nums[i];
while(l < r) {
if((nums[l] + nums[r]) > target || r == i) r--;
else if(nums[l] + nums[r] < target || l == i) l++;
else {
List<Integer> numList = new ArrayList<Integer>();
numList.add(nums[i]);
numList.add(nums[l]);
numList.add(nums[r]);
list.add(numList);
while(l < r && nums[l] == nums[l+1]) l++;
while(l < r && nums[r] == nums[r-1]) r--;
l++;
r--;
}
}
}
return list;
}
}