所以祈祷告诉我，如何从C中的一串垃圾中获取最大的连续字母串？这是一个例子：
char *s = "(2034HEY!!11 th[]thisiswhatwewant44";
会回来......
thisiswhatwewant
我前几天在测验中得到了这个......它让我疯狂（仍然是）试图弄明白！
更新：
我的错，伙计，我忘了包含这样一个事实：你可以使用的唯一功能是strlen功能。从而使它变得更难......
答案 0 :(得分：3)
Uae strtok()将您的字符串拆分为令牌，使用所有非字母字符作为分隔符，并找到最长的令牌。
要查找最长的令牌，您需要为令牌组织一些存储空间 - 我会使用链接列表。
就这么简单。
修改
好的，如果strlen()是唯一允许的函数，你可以先找到源字符串的长度，然后遍历它并用NULL替换所有非字母字符 - 基本上就是strtok()确实
然后，您需要第二次浏览已修改的源字符串，一次推进一个令牌，并使用strlen()找到最长的一个。
答案 1 :(得分：1)
这听起来类似于标准的UNIX'字符串'实用程序。
跟踪由NULL终止的最长可打印字符。 遍历字节直到您点击可打印字符。开始计数。如果你击中一个不可打印的角色停止计数并扔掉起点。如果您遇到NULL，请检查当前运行的长度是否大于先前的记录持有者。如果是这样，记录它，并开始寻找下一个字符串。
答案 2 :(得分：1)
与其他许多子串相比，什么定义了“好”子串 - 只是小写字母？ （即没有空格，数字，标点符号，大写字母和＆amp; c）？
无论检查字符是否为“好”的谓词P，通过s对每个字符应用P的单次传递都可以让您轻松识别每个“好字符”的开始和结束“，记住并挑选最长的。在伪代码中：
longest_run_length = 0
longest_run_start = longest_run_end = null
status = bad
for i in (all indices over s):
if P(s[i]): # current char is good
if status == bad: # previous one was bad
current_run_start = current_run_end = i
status = good
else: # previous one was also good
current_run_end = i
else: # current char is bad
if status == good: # previous one was good -> end of run
current_run_length = current_run_end - current_run_start + 1
if current_run_length > longest_run_length:
longest_run_start = current_run_start
longest_run_end = current_run_end
longest_run_length = current_run_length
status = bad
# if a good run ends with end-of-string:
if status == good: # previous one was good -> end of run
current_run_length = current_run_end - current_run_start + 1
if current_run_length > longest_run_length:
longest_run_start = current_run_start
longest_run_end = current_run_end
longest_run_length = current_run_length
答案 3 :(得分：1)
为什么要使用strlen()？
这是我的版本，它不使用任何功能。
#ifdef UNIT_TEST
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#endif
/*
// largest_letter_sequence()
// Returns a pointer to the beginning of the largest letter
// sequence (including trailing characters which are not letters)
// or NULL if no letters are found in s
// Passing NULL in `s` causes undefined behaviour
// If the string has two or more sequences with the same number of letters
// the return value is a pointer to the first sequence.
// The parameter `len`, if not NULL, will have the size of the letter sequence
//
// This function assumes an ASCII-like character set
// ('z' > 'a'; 'z' - 'a' == 25; ('a' <= each of {abc...xyz} <= 'z'))
// and the same for uppercase letters
// Of course, ASCII works for the assumptions :)
*/
const char *largest_letter_sequence(const char *s, size_t *len) {
const char *p = NULL;
const char *pp = NULL;
size_t curlen = 0;
size_t maxlen = 0;
while (*s) {
if ((('a' <= *s) && (*s <= 'z')) || (('A' <= *s) && (*s <= 'Z'))) {
if (p == NULL) p = s;
curlen++;
if (curlen > maxlen) {
maxlen = curlen;
pp = p;
}
} else {
curlen = 0;
p = NULL;
}
s++;
}
if (len != NULL) *len = maxlen;
return pp;
}
#ifdef UNIT_TEST
void fxtest(const char *s) {
char *test;
const char *p;
size_t len;
p = largest_letter_sequence(s, &len);
if (len && (len < 999)) {
test = malloc(len + 1);
if (!test) {
fprintf(stderr, "No memory.\n");
return;
}
strncpy(test, p, len);
test[len] = 0;
printf("%s ==> %s\n", s, test);
free(test);
} else {
if (len == 0) {
printf("no letters found in \"%s\"\n", s);
} else {
fprintf(stderr, "ERROR: string too large\n");
}
}
}
int main(void) {
fxtest("(2034HEY!!11 th[]thisiswhatwewant44");
fxtest("123456789");
fxtest("");
fxtest("aaa%ggg");
return 0;
}
#endif
答案 4 :(得分：0)
首先，定义“string”并定义“garbage”。你认为什么是有效的非垃圾字符串？写下你可以编程的具体定义 - 这就是编写规范的方法。它是一系列字母数字字符吗？它应该以字母而不是数字开头吗？
一旦弄明白，编程就会非常简单。从一个简单的循环方式开始循环“垃圾”，寻找你需要的东西。完成后，查找有用的C库函数（如strtok）以使代码更精简。
答案 5 :(得分：0)
虽然我等你把这个问题发布为我编码的问题。
此代码遍历传递给“最长”函数的字符串，当它找到字母序列中的第一个时，它会设置指向它的指针并开始计算它的长度。如果它是看到的最长的字母序列，它会将另一个指针（“maxStringStart”指针）设置到该序列的开头，直到找到更长的字符。
最后，它为新字符串分配足够的空间并返回指向它的指针。
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int isLetter(char c){
return ( (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z') );
}
char *longest(char *s) {
char *newString = 0;
int maxLength = 0;
char *maxStringStart = 0;
int curLength = 0;
char *curStringStart = 0;
do {
//reset the current string length and skip this
//iteration if it's not a letter
if( ! isLetter(*s)) {
curLength = 0;
continue;
}
//increase the current sequence length. If the length before
//incrementing is zero, then it's the first letter of the sequence:
//set the pointer to the beginning of the sequence of letters
if(curLength++ == 0) curStringStart = s;
//if this is the longest sequence so far, set the
//maxStringStart pointer to the beginning of it
//and start increasing the max length.
if(curLength > maxLength) {
maxStringStart = curStringStart;
maxLength++;
}
} while(*s++);
//return null pointer if there were no letters in the string,
//or if we can't allocate any memory.
if(maxLength == 0) return NULL;
if( ! (newString = malloc(maxLength + 1)) ) return NULL;
//copy the longest string into our newly allocated block of
//memory (see my update for the strlen() only requirement)
//and null-terminate the string by putting 0 at the end of it.
memcpy(newString, maxStringStart, maxLength);
newString[maxLength + 1] = 0;
return newString;
}
int main(int argc, char *argv[]) {
int i;
for(i = 1; i < argc; i++) {
printf("longest all-letter string in argument %d:\n", i);
printf(" argument: \"%s\"\n", argv[i]);
printf(" longest: \"%s\"\n\n", longest(argv[i]));
}
return 0;
}
这是我在简单C中的解决方案，没有任何数据结构。
我可以在我的终端中运行它：
~/c/t $ ./longest "hello there, My name is Carson Myers." "abc123defg4567hijklmnop890"
longest all-letter string in argument 1:
argument: "hello there, My name is Carson Myers."
longest: "Carson"
longest all-letter string in argument 2:
argument: "abc123defg4567hijklmnop890"
longest: "hijklmnop"
~/c/t $
可以轻松地在isLetter()函数中更改构成字母的标准。例如：
return (
(c >= 'a' && c <= 'z') ||
(c >= 'A' && c <= 'Z') ||
(c == '.') ||
(c == ' ') ||
(c == ',') );
将句号，逗号和空格计为“字母”。
根据您的更新：
将memcpy(newString, maxStringStart, maxLength);替换为：
int i;
for(i = 0; i < maxLength; i++)
newString[i] = maxStringStart[i];
然而，使用C标准库可以更容易地解决这个问题：
char *longest(char *s) {
int longest = 0;
int curLength = 0;
char *curString = 0;
char *longestString = 0;
char *tokens = " ,.!?'\"()@$%\r\n;:+-*/\\";
curString = strtok(s, tokens);
do {
curLength = strlen(curString);
if( curLength > longest ) {
longest = curLength;
longestString = curString;
}
} while( curString = strtok(NULL, tokens) );
char *newString = 0;
if( longest == 0 ) return NULL;
if( ! (newString = malloc(longest + 1)) ) return NULL;
strcpy(newString, longestString);
return newString;
}
答案 6 :(得分：0)
另一种变体。
#include <stdio.h>
#include <string.h>
int main(void)
{
char s[] = "(2034HEY!!11 th[]thisiswhatwewant44";
int len = strlen(s);
int i = 0;
int biggest = 0;
char* p = s;
while (p[0])
{
if (!((p[0] >= 'A' && p[0] <= 'Z') || (p[0] >= 'a' && p[0] <= 'z')))
{
p[0] = '\0';
}
p++;
}
for (; i < len; i++)
{
if (s[i] && strlen(&s[i]) > biggest)
{
biggest = strlen(&s[i]);
p = &s[i];
}
}
printf("%s\n", p);
return 0;
}