unordered_map<string, vector<string>>letters{//声母--韵母 组合{"B",{"a","o","i","u","ai","ei","ao","iao","ie","an","ian","en","in","ang","eng","ing"}}, {"P",{"a","o","i","u","ai","ei","ao","iao","ou","ie","an","ian","en","in","ang","...
unordered_map<string, int> um1({ {"apple", 1}, {"lemon", 2}}); unordered_map<string, int> um3(um1.begin(), um1.end()); // 使用迭代器拷贝构造um1容器某段区间的复制品 简单的使用一下: #include <iostream> #include <vector> #include <unordered_map> using namespace std; int ma...
unordered_map<string, vector<string>>letters{//声母--韵母 组合{"B",{"a","o","i","u","ai","ei","ao","iao","ie","an","ian","en","in","ang","eng","ing"}}, {"P",{"a","o","i","u","ai","ei","ao","iao","ou","ie","an","ian","en","in","ang","...
具体而言,假设有问题的地图是 unordered_map<string, double> 。然后,我想获得 --- 的密钥,以及 vector<string> vector<double> 值。
unordered_map<int, string> mymap; (2)使用n个元素构造unordered_map: unordered_map<int, string> mymap = {{1, "one"}, {2, "two"}, {3, "three"}}; (3)使用给定的范围构造unordered_map: 1 vector<pair<int, string>> myVector = {{1, "one"}, {2, "two"}, {3, "three"}};...
#include <iostream> #include <unordered_map> #include <vector> using namespace std; int main() { vector<pair<string, int>> vec = {{"apple", 1}, {"banana", 2}}; unordered_map<string, int> myMap(vec.begin(), vec.end()); // 从 vector 初始化 for (const auto& pair : myMap...
一般来说,像string、vector、list、deque、forward_list等容器,这些容器的底层逻辑机构为线性序列的数据结构,所以这些容器也叫做序列式容器,序列式容器两个位置存储的值之间一般没有紧密的关联关系,如若将其交换,依旧是序列式容器。序列式容器中的元素是按他们在容器中的存储位置保存和访问的。 HZzzzzLu 2024/11/26...
上面的扩容方式,new了多少个节点,就得销毁多少个节点,所以不太好,下面是另一种方式: 重新开一个vector,将旧表里的桶依次取出放到新表对应的位置上,然后销毁旧表的每个位置,最后再进行交换。 插入完整代码如下: 闭散列完整代码: 登录后发表内容
unordered_map<int, string> myMap; for(auto& pair : myMap) { // 使用 pair.first 和 pair.second 访问键值对 } 复制代码避免频繁拷贝:在遍历unordered_map时,如果需要修改值,应该使用引用或指针避免频繁拷贝。unordered_map<int, vector<int>> myMap; for(auto& pair : myMap) { vector<int>& ...
unordered_map<string, vector<string>>::iterator it; vector<string>::iterator strit; for (it = tmap.begin(); it != tmap.end(); it++) { cout << it->first << ":"; for (strit = it->second.begin(); strit != it->second.end(); strit++) ...