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Problem Statement
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
- Time: O(n^2)O(n2)
- Space: O(|\texttt{ans}|)O(∣ans∣)
3Sum Program Solution In C++
class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
if (nums.size() < 3)
return {};
vector<vector<int>> ans;
sort(begin(nums), end(nums));
for (int i = 0; i + 2 < nums.size(); ++i) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
// choose nums[i] as the first num in the triplet,
// and search the remaining nums in [i + 1, n - 1]
int l = i + 1;
int r = nums.size() - 1;
while (l < r) {
const int sum = nums[i] + nums[l] + nums[r];
if (sum == 0) {
ans.push_back({nums[i], nums[l++], nums[r--]});
while (l < r && nums[l] == nums[l - 1])
++l;
while (l < r && nums[r] == nums[r + 1])
--r;
} else if (sum < 0) {
++l;
} else {
--r;
}
}
}
return ans;
}
};
3Sum Program Solution In Java
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
if (nums.length < 3)
return new ArrayList<>();
List<List<Integer>> ans = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i + 2 < nums.length; ++i) {
if (i > 0 && nums[i] == nums[i - 1])
continue;
// choose nums[i] as the first num in the triplet,
// and search the remaining nums in [i + 1, n - 1]
int l = i + 1;
int r = nums.length - 1;
while (l < r) {
final int sum = nums[i] + nums[l] + nums[r];
if (sum == 0) {
ans.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;
} else if (sum < 0) {
++l;
} else {
--r;
}
}
}
return ans;
}
}
3Sum Program Solution In Python
class Solution:
def threeSum(self, nums: List[int]) -> List[List[int]]:
if len(nums) < 3:
return []
ans = []
nums.sort()
for i in range(len(nums) - 2):
if i > 0 and nums[i] == nums[i - 1]:
continue
l = i + 1
r = len(nums) - 1
while l < r:
sum = nums[i] + nums[l] + nums[r]
if sum == 0:
ans.append((nums[i], nums[l], nums[r]))
l += 1
r -= 1
while nums[l] == nums[l - 1] and l < r:
l += 1
while nums[r] == nums[r + 1] and l < r:
r -= 1
elif sum < 0:
l += 1
else:
r -= 1
return ans
