15. 3Sum

难度: Medium

题目描述

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

解题思路

代码实现

解决方案

java

class Solution {
    public List<List<Integer>> threeSum(int[] nums) {

        List<List<Integer>> result = new ArrayList<>();
        Set<String> filter = new HashSet<>();
        int len = nums.length;
        Arrays.sort(nums);
        for (int i = 0; i < len - 2; i++) {
             // 跳过重复的第一个元素
            if (i > 0 && nums[i] == nums[i - 1]) {
                continue;
            }
            int first = nums[i];

            List<List<Integer>> data = twoSum(nums, 0 - first, i + 1);

            for (List<Integer> d : data) {
                d.add(first);
                d.sort(null);
                String uniqStr = d.toString();
                if (!filter.contains(uniqStr)) {
                    filter.add(uniqStr);
                    result.add(d);
                }
            }

        }

        return result;

    }

    public List<List<Integer>> twoSum(int[] nums, int target, int firstIndex) {
        List<List<Integer>> result = new ArrayList<>();
        Map<Integer, Integer> hashtable = new HashMap<Integer, Integer>();
        for (int i = firstIndex; i < nums.length; ++i) {
            if (hashtable.containsKey(target - nums[i])) {
                List<Integer> data = new ArrayList<>();
                data.add(nums[hashtable.get(target - nums[i])]);
                data.add(nums[i]);
                result.add(data);
             
            }
             // 避免相同元素重复使用
        hashtable.put(nums[i], i);
        }
        return result;
    }
}