-
Notifications
You must be signed in to change notification settings - Fork 77
/
Copy path15.三数之和.java
75 lines (73 loc) · 1.99 KB
/
15.三数之和.java
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
/*
* @lc app=leetcode.cn id=15 lang=java
*
* [15] 三数之和
*
* https://leetcode-cn.com/problems/3sum/description/
*
* algorithms
* Medium (22.51%)
* Likes: 1015
* Dislikes: 0
* Total Accepted: 58.1K
* Total Submissions: 258.1K
* Testcase Example: '[-1,0,1,2,-1,-4]'
*
* 给定一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0
* ?找出所有满足条件且不重复的三元组。
*
* 注意:答案中不可以包含重复的三元组。
*
* 例如, 给定数组 nums = [-1, 0, 1, 2, -1, -4],
*
* 满足要求的三元组集合为:
* [
* [-1, 0, 1],
* [-1, -1, 2]
* ]
*
*
*/
// @lc code=start
import java.util.*;
class Solution {
public List<List<Integer>> threeSum(int[] nums) {
List<List<Integer>> res = new ArrayList<>();
Arrays.sort(nums);
for (int i = 0; i < nums.length-2; i++) {
if (nums[i] > 0) {
break;
}
if (i>0 && nums[i] == nums[i-1]) {
continue;
}
int j = i+1, k = nums.length -1;
while (j < k) {
int sum = nums[i] + nums[j] + nums[k];
if (sum == 0) {
res.add(new ArrayList<>(Arrays.asList(nums[i], nums[j], nums[k])));
while (j < k && nums[j] == nums[j+1]) {
j++;
}
while (j < k && nums[k] == nums[k-1]) {
k--;
}
j++;
k--;
} else if (sum < 0) {
while (j < k && nums[j] == nums[j+1]) {
j++;
}
j++;
} else {
while (j < k && nums[k] == nums[k-1]) {
k--;
}
k--;
}
}
}
return res;
}
}
// @lc code=end