Reverse在cstring中查找指向第n個字符的指針

Question

如何在cstring/BSTR中查找指向第n個字符的指針？

char * RFindNthOccurrence(char* src, char t, int n) 
{ 
    //for i/p string src = "HI,There,you,All" 
    // and t =',' 
    // n =2 
    //returned pointer should be at ",you,All" in same unchanged string 
} 

我找到了第一個和最後一個出現的搜索，但沒有修改字符串反向查找第n個出現是問題。

Answer 1

這個怎麼樣？

#include <iostream> 

const char * RFindNthOccurrence(const char *s, char c, size_t n) 
{ 
    if (!n) return NULL; 

    while ((n -= *s == c) && *s) ++s; 

    return n == 0 ? s : NULL; 
} 

char * RFindNthOccurrence(char *s, char c, size_t n) 
{ 
    if (!n) return NULL; 

    while ((n -= *s == c) && *s) ++s; 

    return n == 0 ? s : NULL; 
} 

int main() 
{ 
    const char *s1 = "HI,There,you,All"; 

    std::cout << RFindNthOccurrence(s1, ',', 2) << std::endl; 

    char s2[] = "HI,There,you,All"; 

    std::cout << RFindNthOccurrence(s2, ',', 2) << std::endl; 

    return 0; 
} 

程序輸出是

,you,All 
,you,All 

功能的行爲方式相同的標準C函數strchr即它找到終止零字符，但只有在情況下，當n = 1

另一個例子

#include <iostream> 

const char * RFindNthOccurrence(const char *s, char c, size_t n) 
{ 
    if (!n) return NULL; 

    while ((n -= *s == c) && *s) ++s; 

    return n == 0 ? s : NULL; 
} 

char * RFindNthOccurrence(char *s, char c, size_t n) 
{ 
    if (!n) return NULL; 

    while ((n -= *s == c) && *s) ++s; 

    return n == 0 ? s : NULL; 
} 

int main() 
{ 
    const char *s = "HI,There,you,All"; 
    const char *p = s; 

    for (size_t i = 1; p = RFindNthOccurrence(s, ',', i); ++i) 
    { 
     std::cout << i << ": " << p << std::endl; 
    } 

    return 0; 
} 

程序輸出是

1: ,There,you,All 
2: ,you,All 
3: ,All 

同樣你可以使用標準C函數strchr而不寫一個特殊函數。例如

#include <iostream> 
#include <cstring> 

int main() 
{ 
    const char *s = "HI,There,you,All"; 
    const char *p = s; 
    size_t n = 2; 

    while ((p = std::strchr(p, ',')) && --n) ++p; 

    if (n == 0) std::cout << p << std::endl; 

    return 0; 
} 

程序輸出是

,you,All 

如果您需要確實是反向搜索則該功能可以像在這個示範項目

#include <iostream> 
#include <cstring> 

const char * RFindNthOccurrence(const char *s, char c, size_t n) 
{ 
    if (!n) return NULL; 

    const char *p = s + std::strlen(s); 

    while ((n -= *p == c) && p != s) --p; 

    return n == 0 ? p : NULL; 
} 

int main() 
{ 
    const char *s = "HI,There,you,All"; 
    const char *p = s; 

    for (size_t i = 1; p = RFindNthOccurrence(s, ',', i); ++i) 
    { 
     std::cout << i << ": " << p << std::endl; 
    } 

    return 0; 
} 

在這種情況下，程序輸出是

1: ,All 
2: ,you,All 
3: ,There,you,All 

Answer 2

這個怎麼樣

// assume n > 0 
char * RFindNthOccurrence(char * str, char t, int n) { 
    int count = 0; 
    char *res = NULL; 
    while (str) { 
     if (t == *str){ 
     ++count; 
     if (count >= n) { 
      res = str; 
     } 
     ++str; 
    } 
    return res; 
}