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 ≤ b ≤ c ≤ 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)
O(n3), add an additional inner loop to perform three sum logic.
public List<List<Integer>> fourSum(int[] num, int target) {
List<List<Integer>> results = new ArrayList<List<Integer>>();
if (num==null||num.length<4) {
return results;
}
Arrays.sort(num);
for (int i=0; i<num.length-3; i++) {
if (i==0 || num[i]>num[i-1]) {
for (int j=i+1; j<num.length-2; j++) {
if (j==i+1 || num[j]>num[j-1]) {
int reverseTarget = target - num[i] - num[j];
int start = j+1;
int end = num.length-1;
while (start<end) {
if (num[start] + num[end] == reverseTarget) {
List<Integer> result = new ArrayList<Integer>();
result.add(num[i]);
result.add(num[j]);
result.add(num[start]);
result.add(num[end]);
start = start + 1;
end = end - 1;
results.add(result);
//compare current start and previous start, if same keep going
while (start<end&&num[start]==num[start-1]) {
start = start + 1;
}
while (start<end&&num[end]==num[end+1]) {
end = end - 1;
}
} else if (num[start] + num[end] < reverseTarget) {
start = start + 1;
} else {
end = end - 1;
}
}
}
}
}
}
return results;
}