15. 3Sum

Posted 2020-11-07 lyinfo

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]

]

 

　　看了半天没啥想法，时间复杂度不要了，暴力解之~

 1     public static List<List<Integer>> threeSum(int[] nums) {
 2         int len = nums.length;
 3         List<List<Integer>> result = new ArrayList<>();
 4         //3个for循环找和为零的组合(要求是n个和就n个嵌套for循环，汗！！！)
 5         for (int i = 0; i < len; i++) {
 6             for (int j = i + 1; j < len; j++) {
 7                 for (int k = len - 1; k > j; k--) {
 8                     if (nums[i] + nums[j] + nums[k] == 0) {
 9                         List<Integer> member = Arrays.asList(nums[i], nums[j], nums[k]);
10                         result.add(member);
11                     }
12                 }
13             }
14         }
15         
16         //获得要去重的元素的索引
17         List<Integer> removeList = new ArrayList<>();
18         for (int i = 0; i < result.size(); i++) {
19             List<Integer> tempi = result.get(i);
20             for (int j = i + 1; j < result.size(); j++) {
21                 List<Integer> tempj = result.get(j);
22                 Collections.sort(tempi);
23                 Collections.sort(tempj);
24                 for (int k = 0; k < 3; k++) {
25                     if (tempi.get(k) != tempj.get(k)) {
26                         break;
27                     }
28                     if (k == 2) {
29                         removeList.add(i);
30                     }
31                 }
32             }
33         }
34 
35         //直接remove会引起索引值变化
36         for (int i = 0; i < result.size(); i++) {
37             for (Integer integer : removeList) {
38                 if (i == integer) {
39                     result.set(i, null);
40                 }
41             }
42         }
43         for (Iterator<List<Integer>> iterator = result.iterator(); iterator.hasNext();) {
44             List<Integer> list = (List<Integer>) iterator.next();
45             if (list == null) {
46                 iterator.remove();
47             }
48         }
49         return result;
50     }

　　测试效果

public static void main(String[] args) {
        int[] nums = { -1, 0, 1, 2, -1, -4 };
        List<List<Integer>> result = threeSum(nums);
        for (List<Integer> list : result) {
            System.out.println(list.toString());
        }
    }

　　结果貌似也是对的，但是以我对leetcode的了解，必然timeOut，提交试试

　　内心毫无波澜，百度之，看了一圈发现，可以从1. Two Sum问题入手，通过排序优化数值查找过程，继续。