给定字符串str，任务是检查字符串是否是 pangram 或不在 C++ 中使用。
A string is a Pangram if the string contains all the English alphabet letters.
例子：
Input: str = “We promptly judged antique ivory buckles for the next prize”Output: YesExplanations: In the above string, str has all the English alphabet letters.
Input: str = “We promptly judged antique ivory buckles for the prize”Output: No
方法 1：不使用 STL
这种方法基于哈希。
- 创建了一个大小为 26 的 boolean 类型的哈希数据结构，索引 0 代表字符’a’，1 代表字符’b’，依此类推。
- 由字符遍历字符串字符和记号特定的字符作为存在于哈希。
- 完成字符串的遍历和标记后，遍历Hash，查看是否所有字符都存在，即每个索引都为真。如果全部被标记，则返回真，否则返回假。
下面是上述方法的实现：
C++
// C++ Program to check if the given
// string is a pangram or not
#include
using namespace std;
// Returns true if the string is
// pangram else false
bool checkPangram(string& str)
{
// Create a hash table to mark
// the characters
// present in the string
vector mark(26, false);
// For indexing in mark[]
int index;
// Traverse all characters
for (int i = 0; i < str.length(); i++) {
// If uppercase character,
// subtract 'A' to find index.
if ('A' <= str[i] && str[i] <= 'Z')
index = str[i] - 'A';
// If lowercase character,
// subtract 'a' to find index.
else if ('a' <= str[i]
&& str[i] <= 'z')
index = str[i] - 'a';
// If this character is not
// an alphabet, skip to next one.
else
continue;
mark[index] = true;
}
// Return false
// if any character is unmarked
for (int i = 0; i <= 25; i++)
if (mark[i] == false)
return (false);
// If all characters were present
return (true);
}
// Driver Code
int main()
{
string str = "We promptly judged"
" antique ivory"
" buckles for the next prize";
if (checkPangram(str) == true)
printf("Yes");
else
printf("No");
return (0);
} CPP
// C++ Program to check whether
// a string pangram or not using STL
#include
using namespace std;
// Function to return given string
// str is pangrams yes or no
string pangrams(string s)
{
// Initialization of count
int count = 0;
// Convert each letter into
// uppercase to avoid counting
// of both uppercase and
// lowercase as different letters
transform(s.begin(),
s.end(),
s.begin(),
::toupper);
// Sort the string
sort(s.begin(), s.end());
// Count distinct alphabets
for (int i = 0; i < s.size(); i++) {
if (s[i] != s[i + 1])
count++;
}
// If count is 27 then the string
// contains all the alphabets
// including space as a
// distinct character
if (count == 27)
return "Yes";
else
return "No";
}
// Driver code
int main()
{
// Given string str
string str = "We promptly "
"judged antique"
"ivory buckles for "
"the next prize";
// Function Call
cout << pangrams(str);
return 0;
} Yes时间复杂度： O(N)，其中 N 是字符串的长度。
辅助空间： O(1)
方法 2：使用STL
STL 的 transform() 方法可用于检查给定的字符串是否为 Pangram。
句法：
transform(s.begin(), s.end(), s.begin(), ::toupper);
方法：
为了检查字符串包含英文字母的所有字母：
Step 1: Firstly convert all the letters into uppercase or lowercase because if it will check without converting, the lowercase and uppercase alphabets will be considered as different letters.
Step 2: Sort the string and check the distinct letter.
Step 3: Space will also be considered as a distinct entity.
Step 4: Now check if count =27 then the string contains all the 26 alphabets.
下面是上述方法的实现：
CPP
// C++ Program to check whether
// a string pangram or not using STL
#include
using namespace std;
// Function to return given string
// str is pangrams yes or no
string pangrams(string s)
{
// Initialization of count
int count = 0;
// Convert each letter into
// uppercase to avoid counting
// of both uppercase and
// lowercase as different letters
transform(s.begin(),
s.end(),
s.begin(),
::toupper);
// Sort the string
sort(s.begin(), s.end());
// Count distinct alphabets
for (int i = 0; i < s.size(); i++) {
if (s[i] != s[i + 1])
count++;
}
// If count is 27 then the string
// contains all the alphabets
// including space as a
// distinct character
if (count == 27)
return "Yes";
else
return "No";
}
// Driver code
int main()
{
// Given string str
string str = "We promptly "
"judged antique"
"ivory buckles for "
"the next prize";
// Function Call
cout << pangrams(str);
return 0;
}
Yes时间复杂度： O(N)，其中 N 是字符串的长度。
辅助空间： O(1)