Leetcode 454. 4Sum II

Given four lists A, B, C, D of integer values, compute how many tuples (i, j, k, l) there are such that A[i] + B[j] + C[k] + D[l] is zero.

To make problem a bit easier, all A, B, C, D have same length of N where 0 ≤ N ≤ 500. All integers are in the range of -228 to 228 - 1 and the result is guaranteed to be at most 231 - 1.

Example:

Input:
A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]

Output:
2

Explanation:
The two tuples are:

1. (0, 0, 0, 1) -> A[0] + B[0] + C[0] + D[1] = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A[1] + B[1] + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

method 1 转化为3Sum

做过前面的题的思想就会想到将4Sum转化为3Sum，但这个方法会TLE

int fourSumCount3(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
	int count = 0;
	unordered_map<int, vector<int>> map;
	for (int i = 0; i < D.size(); i++){
		vector<int> v;
		if (!map.count(D[i])){
			v.push_back(i);
			map[D[i]] = v;
		}
		else{
			v = map[D[i]];
			v.push_back(i);
			map[D[i]] = v;
		}

	}
	for (int i = 0; i < A.size(); i++){
		for (int j = 0; j < B.size(); j++){
			for (int k = 0; k < C.size(); k++){
				int sum3 = A[i] + B[j] + C[k];
				if (map.count(-sum3)){
					vector<int> index = map[-sum3];
					count += index.size();
				}
			}
		}
	}

	return count;
}

method 2 合并数组转化为2Sum

因为本题只需要返回count，而不需要返回具体每个数组相应的索引，因此我们可以将4个数组合并为两个数组，同时记录相同值的个数，然后4Sum就会转化为2Sum。

int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
	unordered_map<int, int> map1;
	unordered_map<int, int> map2;

	for (int i = 0; i < A.size(); i++)
	{
		for (int j = 0; j < B.size(); j++)
		{
			//for map1
			int sumAB = A[i] + B[j];
			if (!map1.count(sumAB)){
				map1[sumAB] = 1;
			}
			else map1[sumAB]++;

			//for map2
			int sumCD = C[i] + D[j];
			if (!map2.count(sumCD)){
				map2[sumCD] = 1;
			}
			else map2[sumCD]++;
		}
	}

	int count = 0;
	unordered_map<int, int>::iterator iter = map1.begin();
	for (; iter != map1.end(); iter++)
	{
		int sumAB = iter->first;
		if (map2.count(-sumAB)) count += (iter->second * map2[-sumAB]);
	}

	return count;
}

method 3 binary Search

本题如果要使用二分查找，也是同样的思路，在ABsum的vector遍历，然后再CDsum的vector中二分查找，只是要考虑下如何再二分查找中统计相同的所有个数
和2Sum中的binary Search差不多

summary

1. 用binary Search在一个有序序列中查找某一个数