4 Sum
Question
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] ]
Solution
这道题和2sum,3sum一样,也是用two pointers的方法。
固定第一个数,移动第二个数(第一个数后面),在第二个数之后的数之中用2sum寻找target。有几点要注意:
注意有重复的数,为了避免重复答案,只能取第一个重复的数
注意conor case,数组长度小于4不行,排序后若最小值的4倍大于targte或者最大值的4倍小于target不行
代码如下:
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> res = new ArrayList<List<Integer>>();
if(nums == null || nums.length < 4){
return res;
}
Arrays.sort(nums);
if(4 * nums[0] > target || 4 * nums[nums.length - 1] < target){
return res;
}
for(int i = 0; i < nums.length - 3; i++){
//skip duplicate
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 head = j + 1;
int tail = nums.length - 1;
while(head < tail){
int sum = nums[i] + nums[j] + nums[head] + nums[tail];
if(sum == target){
List<Integer> list = new ArrayList<Integer>();
list.add(nums[i]);
list.add(nums[j]);
list.add(nums[head]);
list.add(nums[tail]);
res.add(list);
head++;
tail--;
while(head < tail && nums[head] == nums[head - 1]){
head++;
}
while(head < tail && nums[tail] == nums[tail + 1]){
tail--;
}
}else if(sum < target){
head++;
}else{
tail--;
}
}
}
}
return res;
}
}