10:I[25152,["/_next/static/chunks/660e79e7560b2918.js","/_next/static/chunks/6710f67d945945a1.js","/_next/static/chunks/a3af9a4d2a31cc78.js"],"default"] 11:T550,class Solution:def fourSum(self, nums: List[int], target: int) -> List[List[int]]:nums.sort()res = []n = len(nums)for i in range(n - 3): # Skip duplicates for the first number if i > 0 and nums[i] == nums[i - 1]: continue for j in range(i + 1, n - 2): # Skip duplicates for the second number if j > i + 1 and nums[j] == nums[j - 1]: continue left, right = j + 1, n - 1 while left < right: total = nums[i] + nums[j] + nums[left] + nums[right] if total == target: res.append([nums[i], nums[j], nums[left], nums[right]]) left += 1 right -= 1 # Skip duplicates for the third number while left < right and nums[left] == nums[left - 1]: left += 1 # Skip duplicates for the fourth number while left < right and nums[right] == nums[right + 1]: right -= 1 elif total < target: left += 1 else: right -= 1 return res12:T513,class Solution {public List> fourSum(int[] nums, int target) {List> res = new ArrayList<>();if (nums == null || nums.length < 4) return res;Arrays.sort(nums); int n = nums.length; for (int i = 0; i < n - 3; i++) { if (i > 0 && nums[i] == nums[i - 1]) continue; for (int j = i + 1; j < n - 2; j++) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int left = j + 1; int right = n - 1; while (left < right) { // Use long to prevent integer overflow long sum = (long) nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { res.add(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--; } else if (sum < target) { left++; } else { right--; } } } } return res; }}13:T50b,class Solution {public:vector> fourSum(vector& nums, int target) {vector> res;if (nums.size() < 4) return res;sort(nums.begin(), nums.end()); int n = nums.size(); for (int i = 0; i < n - 3; ++i) { if (i > 0 && nums[i] == nums[i - 1]) continue; for (int j = i + 1; j < n - 2; ++j) { if (j > i + 1 && nums[j] == nums[j - 1]) continue; int left = j + 1; int right = n - 1; while (left < right) { // Use long long to prevent integer overflow long long sum = (long long)nums[i] + nums[j] + nums[left] + nums[right]; if (sum == target) { res.push_back({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; } else if (sum < target) { ++left; } else { --right; } } }

Time Complexity

Space Complexity

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