C++ 中的组移位字符串
c++server side programmingprogramming更新于 2025/4/10 2:37:17
假设我们有一个字符串，我们可以将其每个字母"移位"到其连续字母，因此:"abc"可以更改为"bcd"。我们可以继续执行此操作，形成序列:"abc" -> "bcd" -> ... -> "xyz"。如果我们有一个非空字符串列表，其中只包含小写字母，那么我们必须将属于同一移位序列的所有字符串分组。
因此，如果输入为 ["abc", "bcd", "acef", "xyz", "az", "ba", "a", "z"]，则输出为 [ ["abc","bcd","xyz"], ["az","ba"], ["acef"], ["a","z"] ]
为了解决这个问题，我们将遵循以下步骤 −
定义一个映射m
定义一个二维数组 ret
初始化 i := 0，当 i < 字符串大小时，更新(将 i 增加 1)，执行 −
key := 空字符串
初始化 j := 1，当 j < 字符串[i]的大小时，更新(将 j 增加 1)，执行 −
diff := strings[i, j] - strings[i, j - 1]
if diff < 0, then −
diff := diff + 26
key := key 连接 "#" 将 diff 连接为字符串
在 m[key] 末尾插入 strings[i]
对 m 中的每个元素执行 −
在 ret 末尾插入它的值
(将其增加 1)
return ret
示例
让我们看看下面的实现以便更好地理解 −
#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<vector<auto< > v){
cout << "[";
for(int i = 0; i<v.size(); i++){
cout << "[";
for(int j = 0; j <v[i].size(); j++){
cout << v[i][j] << ", ";
}
cout << "],";
}
cout << "]"<<endl;
}
class Solution {
public:
vector<vector<string>> groupStrings(vector<string<& strings) {
unordered_map<string, vector<string> > m;
vector<vector<string< > ret;
for (int i = 0; i < strings.size(); i++) {
string key = "";
for (int j = 1; j < strings[i].size(); j++) {
int diff = strings[i][j] - strings[i][j - 1];
if (diff < 0)
diff += 26;
key += "#" + to_string(diff);
}
m[key].push_back(strings[i]);
}
unordered_map<string, vector<string< >::iterator it = m.begin();
while (it != m.end()) {
ret.push_back(it->second);
it++;
}
return ret;
}
};
main(){
Solution ob;
vector<string< v = {"abc","bcd","acef","xyz","az","ba","a","z"};
print_vector(ob.groupStrings(v));
}
输入
{"abc","bcd","acef","xyz","az","ba","a","z"}
输出
[[az, ba, ],[a, z, ],[abc, bcd, xyz, ],[acef, ],]
