題目
給定4個等長的整型數(shù)組, 每個數(shù)組取一個數(shù)組結(jié)果為0, 輸出一共有多少種可能.例如A, B, C, D, A[a] + B[b] + C[c] + D[d] = 0.
Input: A = [ 1, 2]
B = [-2,-1]
C = [-1, 2]
D = [ 0, 2]
Output: 2
Explanation: 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
思路
先用map記錄兩個數(shù)組的和, 然后計算剩下兩個數(shù)組和, 以之前的map比較.
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int res = 0;
unordered_map<int, int> ab;
for (int a : A) {
for (int b : B) {
ab[a+b]++;
}
}
for (int c : C) {
for (int d : D) {
if (ab.count(-c-d) > 0) {
res += ab[-c-d];
}
}
}
return res;
}
總結(jié)
本體關(guān)鍵是如何最高效的比較, 主要需要降維, 用空間換時間.
