4Sum

4Sum

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note:

• Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
• The solution set must not contain duplicate quadruplets.

    For example, given array S = {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)

Solution

3Sum扩展

Code

class Solution {
	public:
		vector<vector<int>> fourSum(vector<int> &num, int target)
		{
			vector<vector<int>> result;
			int n = num.size();
			if (n < 4)
				return result;
			sort(num.begin(), num.end());
			for (int i = 0; i < n - 3; ++i) {
				if (i > 0 && num[i] == num[i - 1])
					continue;
				vector<int> n(num.begin() + i + 1, num.end());
				vector<vector<int>> ret = threeSum(n, target - num[i]);
				for (auto v : ret)
				{
					v.insert(v.begin(), num[i]);
					result.push_back(v);
				}
			}
			return result;
		}
		vector<vector<int>> threeSum(vector<int> &num, int target) {
		//	sort(num.begin(), num.end());
			vector<vector<int>> result;
			auto n = num.size();
			size_t i;
			for (i = 0; i < n; ++i) {
				if (i > 0 && num[i] == num[i - 1])
					continue;
				int a = num[i];
				int low = i + 1, high = n - 1;
				while (low < high) {
					int b = num[low];
					int c = num[high];
					if (a + b + c == target) {
						vector<int> v = {a, b, c};
						result.push_back(v);
						while (low + 1 < n && num[low] == num[low + 1]) low++;
						while (high - 1 >= 0 && num[high] == num[high - 1]) high--;
						low++;
						high--;
					} else if (a + b + c > target) {
						while (high - 1 >= 0 && num[high] == num[high - 1]) high--;
						high--;
					} else {
						while (low + 1 < n && num[low] == num[low + 1]) low++;
						low++;
					}
				}
			}
			return result;
		}
};