双指针
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
class Solution {
public List<List<Integer>> fourSum(int[] nums, int target) {
/**
* 在《15. 三数之和》的基础上,再套一层for循环
* 添加了一些剪枝优化,减少判断
*/
List<List<Integer>> list = new ArrayList<>();
if (nums.length < 4){
return list;
}
Arrays.sort(nums);
for (int i = 0; i < nums.length - 3; i++) {
if (i > 0 && nums[i] == nums[i - 1]){
continue;
}
/**
* 剪枝
* 数组是排好序的,如果前四个值大于target,结果肯定大于target,直接退出循环
* 因为四个数相加会整型溢出,所以分成两次加法
*/
if (nums[i] + nums[i + 1] > target - (nums[i + 2] + nums[i + 3])){
break;
}
/**
* 同理,和小于target时,当前的i也不满足条件,继续循环
*/
if (nums[i] + nums[nums.length - 1] < target - (nums[nums.length - 2] + nums[nums.length - 3])){
continue;
}
for (int j = i + 1; j < nums.length - 2; j++) {
/**
* j和i不相邻时如果重复,就略过
*/
if (j > i + 1 && nums[j - 1] == nums[j]){
continue;
}
/**
* 剪枝
*/
if (nums[i] + nums[j] > target - (nums[j + 1] + nums[j + 2])){
break;
}
if (nums[i] + nums[j] < target - (nums[nums.length - 1] + nums[nums.length - 2])){
continue;
}
int left = j + 1;
int right = nums.length - 1;
while (left < right){
if (nums[i] + nums[j] > target - (nums[left] + nums[right])){
right--;
}
else if (nums[i] + nums[j] < target - (nums[left] + nums[right])){
left++;
}
else {
list.add(new ArrayList<>(Arrays.asList(nums[i], nums[j], nums[left], nums[right])));
left++;
right--;
while (left < right && nums[left] == nums[left - 1]){
left++;
}
while (left < right && nums[right] == nums[right + 1]){
right--;
}
}
}
}
}
return list;
}
}
/**
* 时间复杂度 O(n^3)
* 空间复杂度 O(1)
*/
https://leetcode-cn.com/problems/4sum/