标签：div first -o ant equal vector ice its not
Given two arrays A and B of equal size, the advantage of A with respect to B is the number of indices i for which A[i] > B[i].
Return any permutation of A that maximizes its advantage with respect to B.
Example 1:
Input: A = [2,7,11,15], B = [1,10,4,11]
Output: [2,11,7,15]
Example 2:
Input: A = [12,24,8,32], B = [13,25,32,11]
Output: [24,32,8,12]
Note:
1 <= A.length = B.length <= 100000 <= A[i] <= 10^90 <= B[i] <= 10^9
Approach #1: C++.
class Solution {
public:
vector<int> advantageCount(vector<int>& A, vector<int>& B) {
map<int, int> m;
for (int i : A) m[i]++;
map<int, int>::iterator it;
vector<int> res;
for (int i : B) {
it = m.upper_bound(i);
int x = it != m.end() ? it->first : m.begin()->first;
if (--m[x] == 0) m.erase(x);
res.push_back(x);
}
return res;
}
};
标签：div first -o ant equal vector ice its not
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10311985.html
