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: 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] ]
通用解决办法
int len = 0;
public List<List<Integer>> fourSum(int[] nums, int target) {
len = nums.length;
Arrays.sort(nums);
return kSum(nums, target, 4, 0);
}
private ArrayList<List<Integer>> kSum(int[] nums, int target, int k, int index) {
ArrayList<List<Integer>> res = new ArrayList<List<Integer>>();
if(index >= len) {
return res;
}
if(k == 2) {
int i = index, j = len - 1;
while(i < j) {
//find a pair
if(target - nums[i] == nums[j]) {
List<Integer> temp = new ArrayList<>();
temp.add(nums[i]);
temp.add(target-nums[i]);
res.add(temp);
//skip duplication
while(i<j && nums[i]==nums[i+1]) {
i++;
}
while(i<j && nums[j-1]==nums[j]) {
j--;
}
i++;
j--;
//move left bound
} else if (target - nums[i] > nums[j]) {
i++;
//move right bound
} else {
j--;
}
}
} else{
for (int i = index; i < len - k + 1; i++) {
//use current number to reduce ksum into k-1sum
ArrayList<List<Integer>> temp = kSum(nums, target - nums[i], k-1, i+1);
if(temp != null){
//add previous results
for (List<Integer> t : temp) {
t.add(0, nums[i]);
}
res.addAll(temp);
}
while (i < len-1 && nums[i] == nums[i+1]) {
//skip duplicated numbers
i++;
}
}
}
return res;
}