LeetCode 3sum 问题

1、

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 ≤ b ≤ 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)

2、

这道题引申与2sum问题，之后还可以继续引申4sum。

具体的思路就是双指针的灵活应用；

3、我的解法：

 1 List<List<Integer>> ret = new ArrayList<List<Integer>>();
 2     public List<List<Integer>> threeSum(int[] num) {
 3
 4         if (num == null || num.length < 3) return ret;
 5
 6         Arrays.sort(num);
 7
 8         int len = num.length;
 9         for (int i = 0; i < len-2; i++) {
10             if (i > 0 && num[i] == num[i-1]) continue;
11             find(num, i+1, len-1, num[i]); //寻找两个数与num[i]的和为0
12         }
13
14         return ret;
15     }
16     public void find(int[] num, int begin, int end, int target) {
17         int l = begin, r = end;
18         while (l < r) {
19             if (num[l] + num[r] + target == 0) {
20                 List<Integer> ans = new ArrayList<Integer>();
21                 ans.add(target);
22                 ans.add(num[l]);
23                 ans.add(num[r]);
24                 ret.add(ans); //放入结果集中
25                 while (l < r && num[l] == num[l+1]) l++;
26                 while (l < r && num[r] == num[r-1]) r--;
27                 l++;
28                 r--;
29             } else if (num[l] + num[r] + target < 0) {
30                 l++;
31             } else {
32                 r--;
33             }
34         }
35     }

转载于:https://www.cnblogs.com/winterRel/p/5195874.html

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