程序打印最后10行

给定一个字符串的某些文本行，每行之间用’\ n’字符分隔。打印最后十行。如果行数少于10，则打印所有行。

资料来源：Microsoft专访|套装10

以下是步骤
1)查找最后出现的DELIM或’\ n’
2)将目标位置初始化为最后一次出现的’\ n’并计数为0，然后在count <10时执行以下操作
…… 2.a)找到“ \ n”的下一个实例并更新目标位置
….. 2.b)跳过“ \ n”并增加计数“ \ n”并更新目标位置
3)从目标位置打印子字符串。

C++

// C++ Program to print the last 10 lines. 
// If number of lines is less than 10,
// then print all lines. 
#include 
using namespace std; 
#define DELIM '\n' 
  
/* Function to print last n lines of a given string */
void print_last_lines(char *str, int n) 
{ 
    /* Base case */
    if (n <= 0) 
    return; 
  
    size_t cnt = 0; // To store count of '\n' or DELIM 
    char *target_pos = NULL; // To store the output position in str 
  
    /* Step 1: Find the last occurrence of DELIM or '\n' */
    target_pos = strrchr(str, DELIM); 
  
    /* Error if '\n' is not present at all */
    if (target_pos == NULL) 
    { 
        cout << "ERROR: string doesn't contain '\\n' character\n"; 
        return; 
    } 
  
    /* Step 2: Find the target position from
       where we need to print the string */
    while (cnt < n) 
    { 
        // Step 2.a: Find the next instance of '\n' 
        while (str < target_pos && *target_pos != DELIM) 
            --target_pos; 
  
        /* Step 2.b: skip '\n' and increment count of '\n' */
        if (*target_pos == DELIM) 
            --target_pos, ++cnt; 
  
        /* str < target_pos means str has 
        less than 10 '\n' characters, so break from loop */
        else
            break; 
    } 
  
    /* In while loop, target_pos is decremented 2 times, 
    that's why target_pos + 2 */
    if (str < target_pos) 
        target_pos += 2; 
  
    // Step 3: Print the string from target_pos 
    cout << target_pos << endl; 
} 
  
// Driver Code
int main(void) 
{ 
    char *str1 ="str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7\nstr8\nstr9"
                "\nstr10\nstr11\nstr12\nstr13\nstr14\nstr15\nstr16\nstr17"
                "\nstr18\nstr19\nstr20\nstr21\nstr22\nstr23\nstr24\nstr25"; 
    char *str2 ="str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7"; 
    char *str3 ="\n"; 
    char *str4 = ""; 
  
    print_last_lines(str1, 10); 
    cout << "-----------------\n"; 
  
    print_last_lines(str2, 10); 
    cout << "-----------------\n";
  
    print_last_lines(str3, 10); 
    cout << "-----------------\n";; 
  
    print_last_lines(str4, 10); 
    cout << "-----------------\n";
  
    return 0; 
} 
  
// This is code is contributed by rathbhupendra

C

/* Program to print the last 10 lines. If number of lines is less
   than 10, then print all lines. */
  
#include 
#include 
#define DELIM   '\n'
  
/* Function to print last n lines of a given string */
void print_last_lines(char *str, int n)
{
    /* Base case */
    if (n <= 0)
       return;
  
    size_t cnt  = 0; // To store count of '\n' or DELIM
    char *target_pos   = NULL; // To store the output position in str
  
    /* Step 1: Find the last occurrence of DELIM or '\n' */
    target_pos = strrchr(str, DELIM);
  
    /* Error if '\n' is not present at all */
    if (target_pos == NULL)
    {
        fprintf(stderr, "ERROR: string doesn't contain '\\n' character\n");
        return;
    }
  
    /* Step 2: Find the target position from where we need to print the string */
    while (cnt < n)
    {
        // Step 2.a: Find the next instance of '\n'
        while (str < target_pos && *target_pos != DELIM)
            --target_pos;
  
         /* Step 2.b: skip '\n' and increment count of '\n' */
        if (*target_pos ==  DELIM)
            --target_pos, ++cnt;
  
        /* str < target_pos means str has less than 10 '\n' characters,
           so break from loop */
        else
            break;
    }
  
    /* In while loop, target_pos is decremented 2 times, that's why target_pos + 2 */
    if (str < target_pos)
        target_pos += 2;
  
    // Step 3: Print the string from target_pos
    printf("%s\n", target_pos);
}
  
// Driver program to test above function
int main(void)
{
    char *str1 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7\nstr8\nstr9"
                 "\nstr10\nstr11\nstr12\nstr13\nstr14\nstr15\nstr16\nstr17"
                 "\nstr18\nstr19\nstr20\nstr21\nstr22\nstr23\nstr24\nstr25";
    char *str2 = "str1\nstr2\nstr3\nstr4\nstr5\nstr6\nstr7";
    char *str3 = "\n";
    char *str4 = "";
  
    print_last_lines(str1, 10);
    printf("-----------------\n");
  
    print_last_lines(str2, 10);
    printf("-----------------\n");
  
    print_last_lines(str3, 10);
    printf("-----------------\n");
  
    print_last_lines(str4, 10);
    printf("-----------------\n");
  
    return 0;
}

输出：

str16
str17
str18
str19
str20
str21
str22
str23
str24
str25
-----------------
str1
str2
str3
str4
str5
str6
str7
-----------------

-----------------
ERROR: string doesn't contain '\n' character
-----------------

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