友情支持
如果您觉得这个笔记对您有所帮助，看在D瓜哥码这么多字的辛苦上，请友情支持一下，D瓜哥感激不尽，😜
|
|
有些打赏的朋友希望可以加个好友，欢迎关注D 瓜哥的微信公众号，这样就可以通过公众号的回复直接给我发信息。
公众号的微信号是: jikerizhi。因为众所周知的原因，有时图片加载不出来。 如果图片加载不出来可以直接通过搜索微信号来查找我的公众号。 |
18. 4Sum
Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.
Note:
The solution set must not contain duplicate quadruplets.
Example:
Given array nums = [1, 0, -1, 0, -2, 2], and target = 0. A solution set is: [ [-1, 0, 0, 1], [-2, -1, 1, 2], [-2, 0, 0, 2] ]
思路分析
-
一刷
-
二刷
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
/**
* Runtime: 4 ms, faster than 94.46% of Java online submissions for 4Sum.
*
* Memory Usage: 40.7 MB, less than 52.17% of Java online submissions for 4Sum.
*/
public List<List<Integer>> fourSum(int[] nums, int target) {
int numCount = 4;
if (Objects.isNull(nums) || nums.length < numCount) {
return Collections.emptyList();
}
Arrays.sort(nums);
int length = nums.length;
if (target < numCount * nums[0] || numCount * nums[length - 1] < target) {
return Collections.emptyList();
}
List<List<Integer>> result = new LinkedList<>();
for (int i = 0; i < length - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]) {
continue;
}
for (int j = i + 1; j < length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) {
continue;
}
int twoSumTarget = target - nums[i] - nums[j];
int minTwoSum = nums[j + 1] + nums[j + 2];
int maxTwoSum = nums[length - 1] + nums[length - 2];
if (twoSumTarget < minTwoSum || maxTwoSum < twoSumTarget) {
continue;
}
for (int m = j + 1, n = length - 1; m < n; ) {
int twoSum = nums[m] + nums[n];
if (twoSum < twoSumTarget) {
while (m < n && nums[m] == nums[m + 1]) {
m++;
}
m++;
} else if (twoSumTarget < twoSum) {
while (m < n && nums[n - 1] == nums[n]) {
n--;
}
n--;
} else {
result.add(Arrays.asList(nums[i], nums[j], nums[m], nums[n]));
while (m < n && nums[m] == nums[m + 1]) {
m++;
}
m++;
while (m < n && nums[n - 1] == nums[n]) {
n--;
}
n--;
}
}
}
}
return result;
}
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
/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-08-15 18:37:58
*/
public List<List<Integer>> fourSum(int[] nums, int target) {
if (nums == null || nums.length < 4) {
return Collections.emptyList();
}
Arrays.sort(nums);
return numSum(nums, 0, 4, target);
}
private List<List<Integer>> numSum(int[] nums, int idx, int cnt, long target) {
// 剩余长度不够，则直接返回
if (nums.length - idx < cnt || cnt < 2) {
return new ArrayList<>();
}
List<List<Integer>> result = new ArrayList<>();
if (cnt == 2) {
int l = idx;
int r = nums.length - 1;
while (l < r) {
int sum = nums[l] + nums[r];
int left = nums[l];
int right = nums[r];
if (sum < target) {
while (l < r && nums[l] == left) {
l++;
}
} else if (target < sum) {
while (l < r && nums[r] == right) {
r--;
}
} else {
result.add(new ArrayList<>(Arrays.asList(left, right)));
while (l < r && nums[l] == left) {
l++;
}
while (l < r && nums[r] == right) {
r--;
}
}
}
} else {
for (int i = idx; i < nums.length; i++) {
int num = nums[i];
List<List<Integer>> temp = numSum(nums, i + 1, cnt - 1, target - num);
for (List<Integer> rs : temp) {
rs.addFirst(num);
result.add(rs);
}
while (i < nums.length - 1 && nums[i] == nums[i + 1]) {
i++;
}
}
}
return result;
}