18.四数之和

这个题不是很难，和前面三数之和思路一样，但是限时加溢出直接让我给了，最后到了250+个样例后还是过不了就交给deepseek了（上次三数之和是倒在重复元素处理，www）

#include <vector>
#include <algorithm>

using namespace std;

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> ans;
        int n = nums.size();
        if (n < 4) return ans;
        
        sort(nums.begin(), nums.end());
        
        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;
                
                if ((long)nums[i] + nums[j] + nums[j + 1] + nums[j + 2] > target) break;
                if ((long)nums[i] + nums[j] + nums[n - 1] + nums[n - 2] < target) continue;
                
                int left = j + 1, right = n - 1;
                while (left < right) {
                    long sum = (long)nums[i] + nums[j] + nums[left] + nums[right];
                    if (sum == target) {
                        ans.push_back({nums[i], nums[j], nums[left], nums[right]});
                        while (left < right && nums[left] == nums[left + 1]) ++left;
                        ++left;
                        while (left < right && nums[right] == nums[right - 1]) --right;
                        --right;
                    } else if (sum < target) {
                        ++left;
                    } else {
                        --right;
                    }
                }
            }
        }
        return ans;
    }
};