18. 4Sum

Two Pointers

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
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<>();
Arrays.sort(nums);
int n = nums.length;
for (int i = 0; i < n - 3; i++ ) {
if (i > 0 && nums[i] == nums[i - 1]) continue;
for (int j = i + 1; j < n - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) continue;
int l = j + 1, r = n - 1;
while (l < r) {
long sum = (long) nums[i] + nums[j] + nums[l] + nums[r];

if (sum == target) {
result.add(Arrays.asList(nums[i], nums[j], nums[l], nums[r]));

// 去重
while (l < r && nums[l] == nums[l + 1]) l++;
while (l < r && nums[r] == nums[r - 1]) r--;

l++;
r--;
} else if (sum < target) {
l++;
} else {
r--;
}
}

}
}
return result;


}
}

Remarks:

  1. Extended from No. 15 and No. 16
  2. TC: $O(n^3)$