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, ) - 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 ,会超时。
可以用一个hashmap先缓存两个数的和,最终复杂度。这个策略也适用于 3Sum 。
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 {
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;
}
}
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);
}
value.add(new int[]{i, j});
}
}
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)){
result.add(Arrays.asList(sol));
used.add(key);
}
}
}
}
return result;
}
}