454. 4Sum II

  • 46.7%

https://leetcode.com/problems/4sum-ii/discuss/

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.

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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

方法一:

https://discuss.leetcode.com/topic/78430/concise-c-11-code-beat-99-5

Concise C++ 11 code beat 99.5%

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int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
unordered_map<int, int> abSum;
for(auto a : A) {
for(auto b : B) {
++abSum[a+b];
}
}
int count = 0;
for(auto c : C) {
for(auto d : D) {
auto it = abSum.find(0 - c - d);
if(it != abSum.end()) {
count += it->second;
}
}
}
return count;
}

我的代码实现:

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class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int n = A.size();
unordered_map<int, int> map;
for(auto a:A)
for(auto b:B)
map[a+b]++;
int cnt = 0;
for(auto c:C)
for(auto d:D)
if(map.find(-c-d)!=map.end())
cnt += map[-c-d]; // 是+map的value,不是cnt++
return cnt;
}
};
谢谢你,可爱的朋友。