## 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)


### 代码

// 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<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> 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;
}
};