# [3Sum][title]

## Description

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

**Note:**

The solution set must not contain duplicate triplets.

**Example:**

```
Given array nums = [-1, 0, 1, 2, -1, -4],

A solution set is:
[
  [-1, 0, 1],
  [-1, -1, 2]
]
```

**Tags:** Array, Two Pointers


## 思路

题意是让你从数组中找出所有三个和为 0 的元素构成的非重复序列，这样的话我们可以把数组先做下排序，然后遍历这个排序数组，用两个指针分别指向当前元素的下一个和数组尾部，判断三者的和与 0 的大小来移动两个指针，其中细节操作就是优化和去重。

```java
class Solution {
    public List<List<Integer>> threeSum(int[] nums) {
        List<List<Integer>> list = new ArrayList<>();
        int len = nums.length;
        if (len < 3) return list;
        Arrays.sort(nums);
        int max = nums[len - 1];
        if (max < 0) return list;
        for (int i = 0; i < len - 2; ) {
            if (nums[i] > 0) break;
            if (nums[i] + 2 * max < 0) {
                while (nums[i] == nums[++i] && i < len - 2) ;
                continue;
            }
            int left = i + 1, right = len - 1;
            while (left < right) {
                int sum = nums[i] + nums[left] + nums[right];
                if (sum == 0) {
                    list.add(Arrays.asList(nums[i], nums[left], nums[right]));
                    while (nums[left] == nums[++left] && left < right) ;
                    while (nums[right] == nums[--right] && left < right) ;
                } else if (sum < 0) ++left;
                else --right;
            }
            while (nums[i] == nums[++i] && i < len - 2) ;
        }
        return list;
    }
}
```


## 结语

如果你同我一样热爱数据结构、算法、LeetCode，可以关注我 GitHub 上的 LeetCode 题解：[awesome-java-leetcode][ajl]



[title]: https://leetcode.com/problems/3sum
[ajl]: https://github.com/Blankj/awesome-java-leetcode