## 3Sum

### 描述

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

Note:

* Elements in a triplet (a,b,c) must be in non-descending order. (ie, $$a \leq b \leq c$$)
* The solution set must not contain duplicate triplets.

For example, given array S = {-1 0 1 2 -1 -4}.

A solution set is:


(-1, 0, 1)
(-1, -1, 2)


### 分析

### 代码

{% if book.java %}
java
// 3Sum
// 先排序，然后左右夹逼，注意跳过重复的数
// Time Complexity: O(n^2)，Space Complexity: O(1)
public class Solution {
public List> threeSum(int[] nums) {
List> result = new ArrayList<>();
if (nums.length < 3) return result;
Arrays.sort(nums);
final int target = 0;

for (int i = 0; i < nums.length - 2; ++i) {
if (i > 0 && nums[i] == nums[i-1]) continue;
int j = i+1;
int k = nums.length-1;
while (j < k) {
if (nums[i] + nums[j] + nums[k] < target) {
++j;
while(nums[j] == nums[j-1] && j < k) ++j;
} else if(nums[i] + nums[j] + nums[k] > target) {
--k;
while(nums[k] == nums[k+1] && j < k) --k;
} else {
++j;
--k;
while(nums[j] == nums[j-1] && j < k) ++j;
while(nums[k] == nums[k+1] && j < k) --k;
}
}
}
return result;
}
};

{% endif %}

{% if book.cpp %}
cpp
// 3Sum
// 先排序，然后左右夹逼，注意跳过重复的数
// Time Complexity: O(n^2)，Space Complexity: O(1)
class Solution {
public:
vector> threeSum(vector& nums) {
vector> result;
if (nums.size() < 3) return result;
sort(nums.begin(), nums.end());
const int target = 0;

auto last = nums.end();
for (auto i = nums.begin(); i < last-2; ++i) {
if (i > nums.begin() && *i == *(i-1)) continue;
auto j = i+1;
auto k = last-1;
while (j < k) {
if (*i + *j + *k < target) {
++j;
while(*j == *(j - 1) && j < k) ++j;
} else if (*i + *j + *k > target) {
--k;
while(*k == *(k + 1) && j < k) --k;
} else {
result.push_back({ *i, *j, *k });
++j;
--k;
while(*j == *(j - 1) && j < k) ++j;
while(*k == *(k + 1) && j < k) --k;
}
}
}
return result;
}
};

{% endif %}

### 相关题目

* [Two sum](two-sum.md)
* [3Sum Closest](3sum-closest.md)
* [4Sum](4sum.md)