## 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)


### 左右夹逼

// 4Sum
// 先排序，然后左右夹逼
// Time Complexity: O(n^3)，Space Complexity: O(1)
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> 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 {
++k;
--l;
while(nums[k] == nums[k-1] && k < l) ++k;
while(nums[l] == nums[l+1] && k < l) --l;
}
}
}
}
return result;
}
}


### HashMap 做缓存

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

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

final HashSet<String> 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<int[]> 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)){
}
}
}
}
return result;
}
}