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用来当做string类型的边》像这道题,各个字符串之间是有关系的我们要在各个字符串之间建边,从而得到一种关系这个图该如何建立?用unordered_map<string,vector<string>>mpkey值为string毫无疑问应为可能有如下情况:所以我们要用vector<string>来保存一个节点上的多条边...
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","...
std::vector<int> _data; public: MyClass(){} std::string GetStr() const { std::string str; for (auto x: _data ) str += std::to_string(x); return str; } }; struct MyClassHash { size_t operator()(const MyClass &cla) const { std::string str = cla.GetStr(); return std...
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。
#include<string> #include<map> #include<set> #include<unordered_map> #include<unordered_set> #include<vector> #include<time.h> using namespace std; //无序+去重 void test_unordered_set() { unordered_set<int> us; us.insert(4);
unordered_map是C++中的哈希表,可以在任意类型与类型之间做映射。 基本操作 引用头文件(C++11):#include <unordered_map> 定义:unordered_map<int,int>、unordered_map<string, double> ... …
#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...
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
class Solution { public: vector<vector<string>> groupAnagrams(vector<string>& strs) { unordered_map<string,vector<string>> map; vector<vector<string>> result; for(auto& str:strs) { string key = str; sort(key.begin(),key.end()); map[key].emplace_back(str); } for(auto m:map) ...