leetcode 454. 4Sum II
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1.题目
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.
有四个长度一样的列表A,B,C,D。求一共有多少种(i,j,k,l)的组合使得A[i] + B[j] + C[k] + D[l] =0
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
2.分析
将列表分两拨,A,B元素分别求和,C,D元素分别求和。然后从两个和数组里面找出相加等于0的组合有多少。
借助unordered_map
3.代码
int fourSumCount(vector<int>& A, vector<int>& B, vector<int>& C, vector<int>& D) { unordered_map<int, int> AB; unordered_map<int, int> CD; for (int i = 0; i < A.size(); i++) { for (int j = 0; j < A.size(); j++) { ++AB[A[i] + B[j]]; ++CD[C[i] + D[j]]; } } int ans = 0; for (auto p : AB) { auto it = CD.find(-p.first); if (it != CD.end()) ans += it->second*p.second; } return ans;}
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