例子:
char test1[] = " ";
char test2[] = " hello z";
char test3[] = "hello world ";
char test4[] = "x y z ";
结果:
" "
" olleh z"
"olleh dlrow "
"x y z "
问题:
Reverse every world in a string, ignore the spaces.
The following is my code. The basic idea is to scan the string, when finding a word, then reverse it. The complexity of the algorithm is O(n), where n is the length of the string.
How do verify it? Is there a better solution?
void reverse_word(char* s, char* e)
{
while (s < e)
{
char tmp = *s;
*s = *e;
*e = tmp;
++s;
--e;
}
}
char* word_start_index(char* p)
{
while ((*p != '\0') && (*p == ' '))
{
++p;
}
if (*p == '\0')
return NULL;
else
return p;
}
char* word_end_index(char* p)
{
while ((*p != '\0') && (*p != ' '))
{
++p;
}
return p-1;
}
void reverse_string(char* s)
{
char* s_w = NULL;
char* e_w = NULL;
char* runner = s;
while (*runner != '\0')
{
char* cur_word_s = word_start_index(runner);
if (cur_word_s == NULL)
break;
char* cur_word_e = word_end_index(cur_word_s);
reverse_word(cur_word_s, cur_word_e);
runner = cur_word_e+1;
}
}
最佳答案
您的代码似乎是正确的,但它是普通的 C。在 C++ 中,使用相同的方法,它可能看起来像这样:
#include <iostream>
#include <string>
#include <iterator>
#include <algorithm>
#include <cctype>
int main()
{
std::string str = " cat cow dog wolf lobster";
std::string result;
auto it = str.begin();
while (it != str.end()) {
while (it != str.end() && isspace(*it)) {
result += *it;
++it;
}
auto begin = it;
while (it != str.end() && !isspace(*it)) {
++it;
}
auto end = it;
result.append(std::reverse_iterator<decltype(end)>(end),
std::reverse_iterator<decltype(begin)>(begin));
// if you want to modify original string instead, just do this:
std::reverse(begin, end);
}
std::cout << result <<'\n';
}
关于c++ - 反转字符串中的每个单词(应该处理空格),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/12879348/