[LeetCode] 4Sum II 四數之和之二

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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:
2Explanation:
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

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這道題是之前那道4Sum的延伸，讓我們在四個數組中各取一個數字，使其和為0。那么墜傻的方法就是遍歷所有的情況，時間復雜度為O(n⁴)。但是我們想想既然Two Sum那道都能將時間復雜度縮小一倍，那么這道題我們使用哈希表是否也能將時間復雜度降到O(n²)呢？答案是肯定的，我們如果把A和B的兩兩之和都求出來，在哈希表中建立兩數之和跟其出現次數之間的映射，那么我們再遍歷C和D中任意兩個數之和，我們只要看哈希表存不存在這兩數之和的相反數就行了，參見代碼如下：

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解法一：

class Solution {
public:int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {int res = 0;unordered_map<int, int> m;for (int i = 0; i < A.size(); ++i) {for (int j = 0; j < B.size(); ++j) {++m[A[i] + B[j]];}}for (int i = 0; i < C.size(); ++i) {for (int j = 0; j < D.size(); ++j) {int target = -1 * (C[i] + D[j]);res += m[target];}}return res;}
};

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這種方法用了兩個哈希表分別記錄AB和CB的兩兩之和出現次數，然后遍歷其中一個哈希表，并在另一個哈希表中找和的相反數出現的次數，參見代碼如下：

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解法二：

class Solution {
public:int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {int res = 0, n = A.size();unordered_map<int, int> m1, m2;for (int i = 0; i < n; ++i) {for (int j = 0; j < n; ++j) {++m1[A[i] + B[j]];++m2[C[i] + D[j]];}}for (auto a : m1) res += a.second * m2[-a.first];return res;}
};

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類似題目：

4Sum

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參考資料：

https://discuss.leetcode.com/topic/67593/clean-java-solution-o-n-2

https://discuss.leetcode.com/topic/67729/concise-8-line-c-solution-with-hashmap-simple-and-clean/2

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