问题描述 链接到标题

18.四数之和

解题思路 链接到标题

双指针法，i, j = i + 1, l, r;

注意去重和溢出。

代码 链接到标题 

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        std::sort(nums.begin(), nums.end());
        vector<vector<int>> res;
        if (nums.size() < 4)
            return res;
        for (int i = 0; i < nums.size() - 3; i++) {
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            for (int j = i + 1; j < nums.size() - 2; j++) {
                if (j > i + 1 && nums[j] == nums[j - 1])
                    continue;
                int l = j + 1, r = nums.size() - 1;
                while (l < r) {
                    long long sum = (long long)nums[i] + (long long)nums[j] + (long long)nums[l] + (long long)nums[r];
                    if (sum < target) {
                        l++;
                        while (l < nums.size() - 1 && nums[l] == nums[l - 1])
                            l++;
                    } else if (sum > target) {
                        r--;
                        while (r > 2 && nums[r] == nums[r + 1])
                            r--;
                    } else {
                        res.push_back({nums[i], nums[j], nums[l], nums[r]});
                        l++;
                        while (l < nums.size() - 1 && nums[l] == nums[l - 1])
                            l++;
                        r--;
                        while (r > 2 && nums[r] == nums[r + 1])
                            r--;
                    }
                }
            }
        }
        return res;
    }
};