leetcode 454. 4Sum II

相关问题

1. Two Sum
15. 3Sum
16. 3Sum Closest
18. 4Sum
454. 4Sum II

Discription

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.

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] + D1 = 1 + (-2) + (-1) + 2 = 0
2. (1, 1, 0, 0) -> A1 + B1 + C[0] + D[0] = 2 + (-1) + (-1) + 0 = 0

思路

哈希表。
将A,B,C,D分成{A,B}和{C,D}两组。
分别计算A[i]+B[j]和C[m]+D[n]的所有情况。
从而将原问题转化为 Two Sum。

时间复杂度：O(n2)
空间复杂度：O(n)

代码

class Solution {public:    int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {        unordered_map<int, int> map;        int count = 0;        for (int i = 0; i < A.size(); i++)        {            for (int j = 0; j < B.size(); j++)            {                int sum = A[i] + B[j];                if (map.find(sum) == map.end())                    map[sum] = 1;                else                    map[sum] += 1;            }        }        for (int i = 0; i < C.size(); i++)        {            for (int j = 0; j < D.size(); j++)            {                int sum = C[i] + D[j];                if (map.find(-sum) != map.end())                    count += map[-sum];            }        }        return count;    }};