18. 4Sum

Given an array nums of n integers and an integer target, are there elements a, b, c, and d in nums 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.

Example:

Given array nums = [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]
]

---

给定一个含n个整数的数组nums和一个整数target，判断nums中是否存在是个整数a、b、c、d，它们满足a+b+c+d=target？若存在，找出所有符合条件的整数对。
分析：此题与3Sum没什么区别，只需要在此基础上多加一层for循环即可，代码如下：

    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> rst;
        sort(nums.begin(), nums.end());
        int cnt = nums.size();
        int left = 0, right = 0;
        int diff = 0, diff1 = 0, tmp = 0;
        for (int i = 0; i < cnt - 3; i++)
        {
            if (i > 0 && nums[i] == nums[i - 1])
                continue;
            diff = target - nums[i];
            for (int j = i + 1; j < cnt - 2; j++)
            {
                if (j > i+1 && nums[j] == nums[j - 1])
                    continue;
                diff1 = diff - nums[j];
                left = j + 1;
                right = cnt - 1;
                while (left < right)
                {
                    tmp = nums[left] + nums[right];
                    if (tmp == diff1)
                    {
                        rst.push_back({ nums[i],nums[j],nums[left],nums[right] });
                        while (left < right)
                        {
                            if (nums[left] == nums[left + 1])
                                left++;
                            else
                                break;
                        }
                        while (left < right)
                        {
                            if (nums[right] == nums[right - 1])
                                right--;
                            else
                                break;
                        }
                        left++;
                        right--;
                    }
                    else if (tmp < diff1)
                        left++;
                    else
                        right--;
                }
            }
        }
        return rst;        
    }