https://leetcode.com/problems/3sum
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.
- code
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
Arrays.sort(nums);
List<List<Integer>> res = new ArrayList<>();
for (int i = 0; i < nums.length - 2; i++){
if (i > 0 && nums[i] == nums[i-1]) continue;
int target = -nums[i];
int l = i + 1, r = nums.length - 1;
while (l < r){
if (nums[l] + nums[r] < target) l++;
else if (nums[l] + nums[r] > target) r--;
else {
res.add(Arrays.asList(nums[l], nums[r], nums[i]));
while (l < r && nums[l] == nums[l+1]) l++;
while (l < r && nums[r] == nums[r-1]) r--;
l++;
r--;
}
}
}
return res;
}
}
- code
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
nums.sort()
res = []
for i in range(len(nums)-2):
l, r = i + 1, len(nums) - 1 # notice l = i + 1
if i > 0 and nums[i] == nums[i-1]:
continue
while l < r:
s = nums[l] + nums[i] + nums[r]
if s > 0:
r -= 1
elif s < 0:
l += 1
else:
res.append([nums[i], nums[l], nums[r]])
while l < r and nums[l] == nums[l+1]:
l += 1
while l < r and nums[r] == nums[r-1]:
r -= 1
r -= 1
l += 1
return res