15. 3Sum
1. Description
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.
2. Example
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.
3. Constraints
- 0 <= nums.length <= 3000
- $-10^5$ <= nums[i] <= $10^5$
4. Solutions
Two Pointers
n = nums.size()
Time complexity: O($n^2$)
Space complexity: O(logn)
class Solution {
public:
vector<vector<int>> threeSum(vector<int> &nums) {
const int n = nums.size();
sort(nums.begin(), nums.end());
vector<vector<int>> indexes;
for (int i = 0; i < n - 2 && nums[i] <= 0; ++i) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
int j = i + 1, k = n - 1;
while (j < k) {
long long sum = 1LL * nums[i] + nums[j] + nums[k];
if (sum < 0) {
++j;
} else if (sum == 0) {
indexes.push_back({nums[i], nums[j], nums[k]});
++j;
while (j < k && nums[j] == nums[j - 1]) {
++j;
}
--k;
while (j < k && nums[k] == nums[k + 1]) {
--k;
}
} else {
--k;
}
}
}
return indexes;
}
};