18. 4Sum
Tags: ‘Array’, ‘Hash Table’, ‘Two Pointers’
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] ]
Solution
方法1， 通用k-sum方法 左右夹逼 O(n^3) O(1)
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new LinkedList<>();
Arrays.sort(nums);
if (nums == null || nums.length < 4) return result;
for (int i = 0; i < nums.length - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicates
for (int j = i + 1; j < nums.length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j - 1]) continue; // skip duplicates
// inner loop
int lo = j + 1, hi = nums.length - 1;
while (lo < hi) {
int sum = nums[i] + nums[j] + nums[lo] + nums[hi];
if (sum == target) {
List<Integer> temp = new LinkedList<>(Arrays.asList(nums[i], nums[j], nums[lo], nums[hi]));
result.add(temp);
while (lo < hi && nums[lo] == nums[lo+1]) lo++;
while (lo < hi && nums[hi] == nums[hi-1]) hi--;
lo++;
hi--;
}else if (sum < target) {
while (lo < hi && nums[lo] == nums[lo+1]) lo++;
lo++;
} else {
while (lo < hi && nums[hi] == nums[hi-1]) hi--;
hi--;
}
}
}
}
return result;
}
方法2， 因为返回的是数值而不是index，所以可以用hashmap做缓存 O(n^3) O(n)
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new LinkedList<>();
if (nums == null || nums.length < 4) return result;
Arrays.sort(nums);
Map<Integer, List<int[]>> cache = new HashMap<>();
for (int i = 0; i < nums.length; i++) {
for (int j = i + 1; j < nums.length; j++) {
List<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});
}
}
Set<String> used = new HashSet<>();
for (int i = 0; i < nums.length - 3; i++) {
if (i > 0 && nums[i] == nums[i-1]) continue; // avoid duplicate
for (int j = i+1; j < nums.length - 2; j++) {
if (j > i + 1 && nums[j] == nums[j-1]) continue; // avoid duplicate
List<int[]> list = cache.get(target - nums[i] - nums[j]);
if (list == null) {
continue;
} else {
// Found a match
for (int[] pair: list) {
if (j >= pair[0]) continue; // overlap
Integer[] temp = new Integer[]{nums[i], nums[j], nums[pair[0]], nums[pair[1]]}; // Integer is required
String key = Arrays.toString(temp);
if (!used.contains(key)) {
result.add(Arrays.asList(temp));
used.add(key);
}
}
}
}
}
return result;
}
第一次尝试
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> list = new LinkedList<>();
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 twoSumTarget = target - nums[i] - nums[j];
//The following 3 lines of code to calculate the min and max of twoSum
int minTwoSum = nums[j+1] + nums[j+2];
int maxTwoSum = nums[nums.length - 1] + nums[nums.length - 2];
if(twoSumTarget < minTwoSum || twoSumTarget > maxTwoSum) continue;
for(int m = j+1,n= nums.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(twoSum > twoSumTarget){
while(m < n && nums[n] == nums[n-1]) n--;
n--;
}else{
List<Integer> tempList = new LinkedList<>(Arrays.asList(nums[i],nums[j],nums[m],nums[n]));
list.add(tempList);
while(m < n && nums[m] == nums[m+1]) m++;
m++;
while(m < n && nums[n] == nums[n-1]) n--;
n--;
}
}
}
}
return list;
}
}