# 15. 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.
``````

Constraints:

• `3 <= nums.length <= 3000`
• `-105 <= nums[i] <= 105`

``````class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
vector<vector<int>> res;
sort(nums.begin(), nums.end());
if (nums.empty() || nums.back() < 0 || nums.front() > 0) return {};
for (int k = 0; k < (int)nums.size() - 2; ++k) {
if (nums[k] > 0) break;
if (k > 0 && nums[k] == nums[k - 1]) continue;
int target = 0 - nums[k], i = k + 1, j = (int)nums.size() - 1;
while (i < j) {
if (nums[i] + nums[j] == target) {
res.push_back({nums[k], nums[i], nums[j]});
while (i < j && nums[i] == nums[i + 1]) ++i;
while (i < j && nums[j] == nums[j - 1]) --j;
++i; --j;
} else if (nums[i] + nums[j] < target) ++i;
else --j;
}
}
return res;
}
};
``````

``````class Solution {
public:
vector<vector<int>> threeSum(vector<int>& nums) {
set<vector<int>> res;
sort(nums.begin(), nums.end());
if (nums.empty() || nums.back() < 0 || nums.front() > 0) return {};
for (int k = 0; k < (int)nums.size() - 2; ++k) {
if (nums[k] > 0) break;
int target = 0 - nums[k], i = k + 1, j = (int)nums.size() - 1;
while (i < j) {
if (nums[i] + nums[j] == target) {
res.insert({nums[k], nums[i], nums[j]});
while (i < j && nums[i] == nums[i + 1]) ++i;
while (i < j && nums[j] == nums[j - 1]) --j;
++i; --j;
} else if (nums[i] + nums[j] < target) ++i;
else --j;
}
}
return vector<vector<int>>(res.begin(), res.end());
}
};
``````

Github 同步地址：

https://github.com/grandyang/leetcode/issues/15

Two Sum

3Sum Smaller

3Sum Closest

4Sum

https://leetcode.com/problems/3sum/

https://leetcode.com/problems/3sum/discuss/7380/Concise-O(N2)-Java-solution

https://leetcode.com/problems/3sum/discuss/7373/Share-my-simple-java-solution

http://www.lifeincode.net/programming/leetcode-two-sum-3-sum-3-sum-closest-and-4-sum-java/

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