lintcode-medium-4 Sum

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.

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)

也和3 Sum基本一样。

public class Solution {
/**
* @param numbers : Give an array numbersbers of n integer
* @param target : you need to find four elements that's sum of target
* @return : Find all unique quadruplets in the array which gives the sum of
*           zero.
*/
public ArrayList<ArrayList<Integer>> fourSum(int[] numbers, int target) {
/* your code */
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

if(numbers == null || numbers.length == 0)
return result;

Arrays.sort(numbers);

for(int i = 0; i < numbers.length - 3; i++){
if(i > 0 && numbers[i] == numbers[i - 1])
continue;

for(int j = i + 1; j < numbers.length - 2; j++){
if(j > i + 1 && numbers[j] == numbers[j - 1])
continue;

int left = j + 1;
int right = numbers.length - 1;

while(left < right){
if(numbers[i] + numbers[j] + numbers[left] + numbers[right] == target){
ArrayList<Integer> line = new ArrayList<Integer>();
line.add(numbers[i]);
line.add(numbers[j]);
line.add(numbers[left]);
line.add(numbers[right]);
result.add(new ArrayList<Integer>(line));

left++;
while(left < right && numbers[left] == numbers[left - 1])
left++;

right--;
while(left < right && numbers[right] == numbers[right + 1])
right--;
}
else if(numbers[i] + numbers[j] + numbers[left] + numbers[right] < target){
left++;
while(left < right && numbers[left] == numbers[left - 1])
left++;
}
else{
right--;
while(left < right && numbers[right] == numbers[right + 1])
right--;
}
}
}
}

return result;
}
}