Leetcode454

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 -2^28 to 2^28 - 1 and the result is guaranteed to be at most 2^31 - 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:

#include <iostream>
#include <vector>
#include <unordered_map>

using namespace std;

// 时间复杂度：o（n^2）
// 空间复杂度：o（n^2）
class Solution {
public:
    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {

        unordered_map<int, int> record;
        for ( int i = 0; i < C.size(); i++)
            for ( int j = 0; j < D.size(); j++)
                record[ C[i] + D[j] ] ++;

        int res = 0;
        for ( int i = 0; i < A.size(); i++)
            for ( int j = 0; j < B.size(); j++)
                if ( record.find( 0 - A[i] - B[j] ) != record.end() )
                    res += record[ 0 - A[i] - B[j] ];

        return res;
    }
};

总结： 题目中给定n最大是500，所以一个n^2级别的复杂度就可以解决这个问题了。要学会灵活的选取map中的键值对，此题中是值和频次。暴力解法（4重循环）的复杂度为n^4，不符合题意。