# LeetCode454. 4Sum II（思维）

## LeetCode454. 4Sum II（思维）

• Medium
• Accepted：56,454
• Submissions：113,653

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

### 链接

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

### 题解

• 四重循环暴力，时间复杂度$O(n^4)$
• 将D中元素放入查找表，使用unordered_map，时间复杂度$O(n^3)$
• 将C+D的每一种可能放入查找表，时间复杂度$O(n^2)$

### 代码

• Runtime: 208 ms, faster than 76.60% of C++ online submissions for 4Sum II.
• Memory Usage: 28.8 MB, less than 100.00% of C++ online submissions for 4Sum II.
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
unordered_map <int, int> umap;
for (int i = 0; i < C.size(); ++i)
for (int j = 0; j < D.size(); ++j)
umap[C[i]+D[j]]++;
int ans = 0;
for (int i = 0; i < A.size(); ++i)
{
for (int j = 0; j < B.size(); ++j)
{
int x = -1 * (A[i] + B[j]);
if (umap.count(x) > 0)
ans += umap[x];
}
}
return ans;
}
};
• 因为不存在的数默认的值为0，所以无需判断是否在查找表中，直接加即可，简化代码如下：
class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
unordered_map <int, int> umap;
for (int c : C)
for (int d : D)
umap[c + d]++;
int ans = 0;
for (int a : A)
for (int b : B)
ans += umap[-a-b];
return ans;
}
};

The end.
2019年2月17日 星期日