算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## 4Sum


### 描述

Given an array `S` of `n` integers, are there elements `a, b, c`, and `d` in `S` such that `a + b + c + d = target`? Find all unique quadruplets in the array which gives the sum of target.

Note:

* Elements in a quadruplet `(a,b,c,d)` must be in non-descending order. (ie, $$a \leq b \leq c \leq d$$)
* The solution set must not contain duplicate quadruplets.


For example, given array `S = {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)
```



### 分析

先排序,然后左右夹逼,复杂度 $$O(n^3)$$,会超时。

可以用一个hashmap先缓存两个数的和,最终复杂度$$O(n^3)$$。这个策略也适用于 3Sum 。


### 左右夹逼

{% if book.java %}
```java
// 4Sum
// 先排序,然后左右夹逼
// Time Complexity: O(n^3),Space Complexity: O(1)
public class Solution {
public List> fourSum(int[] nums, int target) {
List> result = new ArrayList<>();
if (nums.length < 4) return result;
Arrays.sort(nums);

for (int i = 0; i < nums.length - 3; ++i) {
if (i > 0 && nums[i] == nums[i-1]) continue;
for (int j = i + 1; j < nums.length - 2; ++j) {
if (j > i+1 && nums[j] == nums[j-1]) continue;
int k = j + 1;
int l = nums.length - 1;
while (k < l) {
final int sum = nums[i] + nums[j] + nums[k] + nums[l];
if (sum < target) {
++k;
while(nums[k] == nums[k-1] && k < l) ++k;
} else if (sum > target) {
--l;
while(nums[l] == nums[l+1] && k < l) --l;
} else {
result.add(Arrays.asList(nums[i], nums[j], nums[k], nums[l]));
++k;
--l;
while(nums[k] == nums[k-1] && k < l) ++k;
while(nums[l] == nums[l+1] && k < l) --l;
}
}
}
}
return result;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// 4Sum
// 先排序,然后左右夹逼
// Time Complexity: O(n^3),Space Complexity: O(1)
class Solution {
public:
vector> fourSum(vector& nums, int target) {
vector> result;
if (nums.size() < 4) return result;
sort(nums.begin(), nums.end());

for (int i = 0; i < nums.size() - 3; ++i) {
if (i > 0 && nums[i] == nums[i-1]) continue;
for (int j = i + 1; j < nums.size() - 2; ++j) {
if (j > i+1 && nums[j] == nums[j-1]) continue;
int k = j + 1;
int l = nums.size() - 1;
while (k < l) {
const int sum = nums[i] + nums[j] + nums[k] + nums[l];
if (sum < target) {
++k;
while(nums[k] == nums[k-1] && k < l) ++k;
} else if (sum > target) {
--l;
while(nums[l] == nums[l+1] && k < l) --l;
} else {
result.push_back({nums[i], nums[j], nums[k], nums[l]});
++k;
--l;
while(nums[k] == nums[k-1] && k < l) ++k;
while(nums[l] == nums[l+1] && k < l) --l;
}
}
}
}
return result;
}
};
```
{% endif %}


### HashMap 做缓存

{% if book.java %}
```java
// 4Sum
// 先排序,然后左右夹逼
// Time Complexity: O(n^3),Space Complexity: O(1)
public class Solution {
public List> fourSum(int[] nums, int target) {
List> result = new ArrayList<>();
if (nums.length < 4) return result;
Arrays.sort(nums);

final HashMap> cache = new HashMap<>();
for (int i = 0; i < nums.length; ++i) {
for (int j = i + 1; j < nums.length; ++j) {
ArrayList value = cache.get(nums[i] + nums[j]);
if (value == null) {
value = new ArrayList<>();
cache.put(nums[i] + nums[j], value);
}
value.add(new int[]{i, j});
}
}

final HashSet used = new HashSet<>(); // avoid duplicates
for (int i = 0; i < nums.length; ++i) {
if (i > 0 && nums[i] == nums[i-1]) continue;
for (int j = i + 1; j < nums.length - 2; ++j) {
if (j > i+1 && nums[j] == nums[j-1]) continue;
final ArrayList list = cache.get(target - nums[i] - nums[j]);
if (list == null) continue;;
for (int[] pair : list) {
if (j >= pair[0]) continue; // overlap

final Integer[] sol = new Integer[]{nums[i], nums[j], nums[pair[0]], nums[pair[1]]};
Arrays.sort(sol);
final String key = Arrays.toString(sol);

if(!used.contains(key)){
result.add(Arrays.asList(sol));
used.add(key);
}
}
}
}
return result;
}
}
```
{% endif %}


{% if book.cpp %}
```cpp
// 4Sum
// 用一个hashmap先缓存两个数的和
// Time Complexity: 平均O(n^2),最坏O(n^4),Space Complexity: O(n^2)
class Solution {
public:
vector > fourSum(vector &nums, int target) {
vector> result;
if (nums.size() < 4) return result;
sort(nums.begin(), nums.end());

unordered_map > > cache;
for (size_t a = 0; a < nums.size(); ++a) {
for (size_t b = a + 1; b < nums.size(); ++b) {
cache[nums[a] + nums[b]].push_back(pair(a, b));
}
}

for (int c = 0; c < nums.size(); ++c) {
for (size_t d = c + 1; d < nums.size(); ++d) {
const int key = target - nums[c] - nums[d];
if (cache.find(key) == cache.end()) continue;

const auto& vec = cache[key];
for (size_t k = 0; k < vec.size(); ++k) {
if (c <= vec[k].second)
continue; // 有重叠

result.push_back( { nums[vec[k].first],
nums[vec[k].second], nums[c], nums[d] });
}
}
}
sort(result.begin(), result.end());
result.erase(unique(result.begin(), result.end()), result.end());
return result;
}
};
```
{% endif %}


### 相关题目


* [Two sum](two-sum.md)
* [3Sum](3sum.md)
* [3Sum Closest](3sum-closest.md)