LeetCode Problem
15. 3Sum
Link to LeetCode
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.
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.
public List< List< Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
ArrayList< List< Integer>> result = new ArrayList<>();
for (int i = 0; i < nums.length; i++) {
int j = i + 1;
int k = nums.length - 1;
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
while (j < k) {
if (k < nums.length - 1 && nums[k] == nums[k + 1]) {
k--;
continue;
}
if (nums[i] + nums[j] + nums[k] > 0) {
k--;
} else if (nums[i] + nums[j] + nums[k] < 0) {
j++;
} else {
ArrayList< Integer> l = new ArrayList<>();
l.add(nums[i]);
l.add(nums[j]);
l.add(nums[k]);
result.add(l);
j++;
k--;
}
}
}
return result;
}