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题目:找出字符串中同时包含大写和小写的字母,并返回字母的个数:
eg: (1)"I Love you" : return 0;
(2)"I Like you" :return 1; (Ii)
(3)"I Like you,Lily" :return 2; (IiLl)
(4)"AaBBBBbCcdddde" :return 3; (AaBbCc)
解法一:当你需要快速的获取对应key的value的时候,就可以使用map了。这里利用map中键的唯一性和和大、小写的映射关系来解题。
int count_of_letter_both_upper_lower_case(const char *str)
{
int a = 0;
int i = 0;
map<char, char> map_chars_find;
map<char, char>::iterator it,pend,index;
while (str[i] != '\0')
{
if ('A' <= str[i] && str[i] <= 'Z' || 'a' <= str[i] && str[i] <= 'z')
{
map_chars_find.insert(pair<char, char>(str[i], ' '));
}
i++;
}
it = map_chars_find.begin();
pend = map_chars_find.end();
for (; it != pend; it++)
{
int aa = it->first +32;
index = map_chars_find.find(aa);
if (index != map_chars_find.end())
{
a++;
}
}
cout << "::" << a << endl;
return a;
}
解法二:用两个长度是26的bool类型数组存储已找到的字符。
int count_of_letter_both_upper_lower_case_bool(const char *str)
{
int letter_count = 0;
bool letter_lower[26] = {0};
bool letter_upper[26] = {0};
int i = 0;
while(str[i] != '\0')
{
if(str[i] >= 'A' && str[i] <= 'Z')
{
letter_upper[str[i] - 65] = true;
}
else if(str[i] >= 'a' && str[i] <= 'z')
{
letter_lower[str[i] - 97] = true;
}
i++;
}
for (int j = 0;j < 26;j++)
{
if (letter_lower[j] == true &&letter_upper[j] == true)
{
letter_count++;
}
}
return letter_count;
}