# [LeetCode] 454. 4Sum II 四数之和之二

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.

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

class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int res = 0;
unordered_map<int, int> m;
for (int i = 0; i < A.size(); ++i) {
for (int j = 0; j < B.size(); ++j) {
++m[A[i] + B[j]];
}
}
for (int i = 0; i < C.size(); ++i) {
for (int j = 0; j < D.size(); ++j) {
int target = -1 * (C[i] + D[j]);
res += m[target];
}
}
return res;
}
};

class Solution {
public:
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) {
int res = 0, n = A.size();
unordered_map<int, int> m1, m2;
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
++m1[A[i] + B[j]];
++m2[C[i] + D[j]];
}
}
for (auto a : m1) res += a.second * m2[-a.first];
return res;
}
};

4Sum

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

https://leetcode.com/problems/4sum-ii/discuss/93920/Clean-java-solution-O(n2)

https://leetcode.com/problems/4sum-ii/discuss/93925/Concise-C%2B%2B-11-code-beat-99.5

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2016-11-17 12:38  Grandyang  阅读(13524)  评论(4编辑  收藏  举报