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leetcode : 4 sum (再次提交，超时)
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] ]
tag : two points
public class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(nums == null || nums.length < 4){
return result;
}
Arrays.sort(nums);
for(int i = 0; i < nums.length - 3; i++){
int a = nums[i];
if(i != 0 && nums[i-1] == nums[i]){
continue;
}
int j;
for(j = i + 1; j < nums.length - 2; j ++){
int b = nums[j];
if(j != i + 1 && nums[j-1] == nums[j] ){
continue;
}
int left = j + 1;
int right = nums.length - 1;
while(left < right){
int sum = a + b + nums[left] + nums[right];
if(sum == target){
List<Integer> list = new ArrayList<Integer>();
list.add(a);
list.add(b);
list.add(nums[left]);
list.add(nums[right]);
result.add(list);
left++;
right--;
while(left < right && nums[left] == nums[left-1]){
left++;
}
while(left < right && nums[right+1] == nums[right]){
right--;
}
}else if(sum < target){
left++;
}else{
right--;
}
}
}
}
return result;
}
}
leetcode : 4 sum (再次提交，超时)
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