题目描述:
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
解题思路:
若是用枚举4个数组的方法,时间复杂度是O(N4),毫无疑问会超时,但若是将A_B和C_D两个数组相加,再进行两个数组的比对查找就可以大大缩减所用时间。
代码:
1 class Solution {
2 public:
3 int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
4 unordered_map<long long, int> A_B;
5 for(int i = 0; i < A.size(); i++){
6 for(int j = 0; j < B.size(); j++){
7 int t = A[i]+B[j];
8 A_B[t]++;
9 }
10 }
11 long long ret = 0;
12 for(int i = 0; i < C.size(); i++){
13 for(int j = 0; j < D.size(); j++){
14 int t = -C[i]-D[j];
15 unordered_map<long long, int>::iterator it = A_B.find(t);
16 if(it != A_B.end())
17 ret += it->second;
18 }
19 }
20 return ret;
21 }
22 };