# [leetcode] 454. 4Sum II 解题报告

2016-11-23 09:24:00      个评论    来源：小榕流光的专栏

Given four lists A, B, C, D of integer values, compute how many tuples`(i, j, k, l)there are such thatA[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 -228to 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``````

`思路：一种最brute force的方法是四重循环， 时间复杂度为O(n^4). 稍微优化一下将D数组的元素放到hash表中，第四个元素的查找即可达到O(1)，这样总的时间复杂度为O(n^3), 空间复杂度为O(n)。再进一步可以两个两个的求元素和，也就是先求出A,B数组的任意两个元素和放到hash表中，然后在计算C, D任意两个元素和的时候查找一下hash看能不能找到C, D这两个元素的负数，如果可以找到说明这四个元素相加是为0的。还需要注意的重复元素，在做hash的时候value为个数即可。这样时间复杂度降为O(n^2), 空间复杂度为O(n^2).`

`代码如下：`

``````class Solution {
public:
int fourSumCount(vector& A, vector& B, vector& C, vector& D) {
unordered_map hash1, hash2;
int ans = 0;
for(int i = 0; i < A.size(); i++)
for(int j = 0; j < B.size(); j++)
hash1[A[i]+B[j]]++;
for(int i = 0; i < C.size(); i++)
{
for(int j = 0; j < D.size(); j++)
ans += hash1.count(-C[i] - D[j])? hash1[-C[i] - D[j]]: 0;
}
return ans;
}
};``````