# 18 4Sum – Medium

### Problem:

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)

### Thoughts:

This is very similar to 3Sum and 4Sum problem. Just increase one level of for loop.

### Solutions:

public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
if (nums.length < 4) {
return result;
}
Arrays.sort(nums);
int max = nums[nums.length -1];
for (int i = 0; i < nums.length; i++) {
if (i > 0 && nums[i] == nums[i-1]) {
continue;
}
if (nums[i] + 3 * max < target) {
continue;// nums[i] is too small
}
if (nums[i] * 4 > target) {
break;//nums[i] is too bigger
}
for (int j = i + 1; j < nums.length; j++) {
if (j > i + 1 && nums[j] == nums[j-1]) {
continue;
}
int start = j + 1;
int end = nums.length - 1;
while (start < end) {
int sum = nums[i] + nums[j] + nums[start] + nums[end];
if (sum == target) {
int startVal = nums[start];
int endVal = nums[end];
while (start < end && startVal == nums[start]) {
start ++;
}
while (start < end && endVal == nums[end]) {
end --;
}
}
else if (sum < target) {
start ++;
}
else {
end --;
}
}
}
}
return result;
}
}