[LeetCode] 454. 4Sum II

轉自LeetCode

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

<Solution>
這題最暴力的解法，就是找出所有和，這樣時間複雜度會是 O(n^4)

使用 hash map 來做簡化到 O(n^2)

• 用兩個 hash map 來存 A+B 和 C+D 的所有情況

• 如果 A+B+C+D 可以等於 0，那就代表 A+B = -(C+D)。因此，可以用 map[-(A+B)] 來看是不是存在相對應的 C+D

code 如下(參考資料)

C++

	class Solution {
	public:
	int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
	int ans = 0;
	unordered_map<int, int> map;
	const int len = A.size();
	for(int i = 0; i < len; i++) {
	for(int j = 0; j < len; j++) {
	map[A[i] + B[j]]++;
	}
	}
	int sum;
	for(int i = 0; i < len; i++) {
	for(int j = 0; j < len; j++) {
	sum = -(C[i]+D[j]);
	ans += (map.find(sum) != map.end()) ? map[sum] : 0;
	}
	}
	return ans;
	}
	};

view raw fourSumII.cpp hosted with ❤ by GitHub

Java

	public class Solution {
	public int fourSumCount(int[] A, int[] B, int[] C, int[] D) {
	int ans = 0;
	HashMap<Integer, Integer> hashMap = new HashMap<>();
	final int len = A.length;
	int sum;
	//>> calculate all combinations of A and B
	for(int i = 0; i < len; i++) {
	for(int j = 0; j < len; j++) {
	sum = A[i] + B[j];
	hashMap.put(sum, hashMap.getOrDefault(sum, 0) + 1);
	}
	}
	//>> find the counter part of A+B
	for(int i = 0; i < len; i++) {
	for(int j = 0; j < len; j++) {
	ans += hashMap.getOrDefault(-1 * (C[i] + D[j]), 0);
	}
	}
	return ans;
	}
	}

view raw fourSumII.java hosted with ❤ by GitHub