我正在练习我的C& C ++技能,然后我决定使用两种语言中使用的方法来解决字符串反向问题。我写了一个递归解决方案和索引方法。这里有4个反向功能; 2使用严格的C方法进行计算,另外2使用C ++(STL,String,algorithm)调用。
// C++ reverse string
#include <string> // string
#include <algorithm> // reverse
#include <iostream> // cout
#include <cstring> // std::strcpy
#include <stdio.h> // printf
#include <sys/time.h> // gettimeofday
inline void swap_characters(char* left, char* right) {
char temp = *left;
*left = *right;
*right = temp;
}
void c_index_reverse(char* input, size_t inputSize) {
const size_t strSize = inputSize - 1;
char temp;
for(int i=0 ; i < inputSize / 2 ; i++) {
swap_characters(&input[i], &input[strSize - i]);
}
}
void c_recursive_reverse(char str[], int index, int size)
{
swap_characters(&str[index], &str[size - index]);
if (index == size / 2)
return;
c_recursive_reverse(str, index + 1, size);
}
void c_plusplus_index_reverse(std::string& input) {
const size_t strSize = input.length();
for(int i=0 ; i < strSize / 2 ; i++)
std::swap(input[i], input[strSize - i - 1]);
}
std::string c_plusplus_recursive_reverse(std::string& input) {
if(input.length() <= 1) {
return input;
}
std::string tmp = std::string(input.begin() + 1, input.end());
return c_plusplus_recursive_reverse(tmp) + input[0];
}
double timeit(struct timeval &start, struct timeval &end){
double delta = ((end.tv_sec - start.tv_sec) * 1000000u + end.tv_usec - start.tv_usec) / 1.e6;
return delta;
}
int main() {
struct timeval start,end;
// using C++ STL
std::string temp = "something very weird is another word that includes a longer text to see the delay" \
"something very weird is another word that includes a longer text to see the delay" \
"something very weird is another word that includes a longer text to see the delay" \
"something very weird is another word that includes a longer text to see the delay" \
"something very weird is another word that includes a longer text to see the delay" \
"something very weird is another word that includes a longer text to see the delay" \
"something very weird is another word that includes a longer text to see the delay";
std::cout << temp << std::endl;
// using c++ recursive reverse function - 4
gettimeofday(&start, NULL);
std::reverse(temp.begin(), temp.end());
gettimeofday(&end, NULL);
std::cout << temp << std::endl;
printf("%lf \n",timeit(start, end));
// use C++ style functions
// using recersive - 5
gettimeofday(&start, NULL);
temp = c_plusplus_recursive_reverse(temp);
gettimeofday(&end, NULL);
std::cout << temp << std::endl;
printf("%lf \n",timeit(start, end));
// using index reverse - 3
gettimeofday(&start, NULL);
c_plusplus_index_reverse(temp);
gettimeofday(&end, NULL);
std::cout << temp << std::endl;
printf("%lf \n",timeit(start, end));
// Now do C style
char *cStr = new char[temp.length() + 1];
std::strcpy(cStr, temp.c_str());
// using index - 1
gettimeofday(&start, NULL);
c_index_reverse(cStr, temp.length());
gettimeofday(&end, NULL);
printf("%s \n", cStr);
printf("%lf \n",timeit(start, end));
// using recersive - 2
gettimeofday(&start, NULL);
c_recursive_reverse(cStr, 0, temp.length() - 1);
gettimeofday(&end, NULL);
printf("%s \n", cStr);
printf("%lf \n",timeit(start, end));
return 0;
}
答案 0 :(得分:0)
使用内联函数swap_characters()
只能在三行代码中做大工作,在已有代码时创建更多指针(尽管编译器可能会优化)。使用另一个索引变量,比如说j
,它会减少直到遇到i
会更有效。
void c_index_reverse(char* input, size_t inputSize) {
int j = inputSize - 1;
char temp;
for(int i=0; i<j; i++) {
temp = input[i];
input[i] = input[j];
input[j--] = temp;
}
}
“我还想知道每种方法使用多少内存”。非递归方法使用最少的内存,因为它只使用指针和索引器。但是递归方法使用了更多的内存,因为每个字符串字符(最多半个字符串长度)都会调用递归,因此堆栈的使用会更多。由于你的字符串"something very weird ..."
有大约600个字符,这就是大量的堆栈使用,并且大部分执行时间都花在调用,操作堆栈帧和返回上,而花费很少的时间来交换字符。
这里的递归是“隐藏在无处”。
答案 1 :(得分:0)
C ++递归函数非常糟糕:使用迭代器可以提高速度并且代码更清晰:
g++ test.cpp -lmpfr -lgmp