C语言(函数)学习之[strstr]&[strcasestr]一.strstr函数使用[1]函数原型char*strstr(constchar*haystack,constchar*needle);[2]头文件#include <string.h>[3]函数功能搜索"子串"在"指定字符串"中第一次出现的位置{4}参数说明haystack-->被查找的目标字符串"父串"needle-->要查找的字符串对象"

我在github有对PHP源码更详细的注解.感兴趣的可以围观一下,给个star.PHP5.4源码注解.可以通过commit记录查看已添加的注解. strpos mixed strpos ( string $haystack, mixed $needle [, int $offset = 0 ] ) 如果offset指定了,查找会从offset的位置开始.offset不能为负数. 返回needle第一次出现在haystack的位置.如果在haystack中找不到needle,则返回FALSE. n

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Update (2014-11-02): The signature of the function had been updated to return the index instead of the pointer. If you still

strstr函数:返回主串中子字符串的位置后的所有字符. #include <stdio.h> const char *my_strstr(const char *str, const char *sub_str) { ; str[i] != '\0'; i++) { int tem = i; //tem保留主串中的起始判断下标位置 ; while(str[i++] == sub_str[j++]) { if(sub_str[j] == '\0') { return &str[tem]

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 1 class Solution(object): def strStr(self, haystack, needle): """ :type haystack: str :type needle: str :rtype: in

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 注:直接看题目可能会有些偏差,因为容易认为 needle是一个字符,那就也太容易了,实则,haystack和needle都是字符串(注意是字符串不是数组) 解法思路: 暴力搜索: public class Solution { public int

1. 讨论目标字符串若为空, 则返回-1: 资源字符串若为空, 则返回-1. 2.讨论目标字符串个数为零, 则返回0: 资源字符串个数为零, 则返回-1. 3. 插入旗帜来使第二循环的结束为有条件地返回(为true才返回, 为false则break跳到上循环继续). class Solution { /** * Returns a index to the first occurrence of target in source, * or -1 if target is not part of

原型:char * strstr( char *haystack,  char *needle ) 用法:#include <string.h> 功能:在haystack中寻找needle第一次出现的位置(不比较结束字符NULL). 说明:返回指向第一次出现的needle位置的指针,如果没有找到则返回NULL.

C语言函数 编辑 包含文件:string.h 函数名: strstr 函数原型:      extern char *strstr(char *str1, const char *str2); 语法: * strstr(str1,str2) str1: 被查找目标 string expression to search. str2: 要查找对象 The string expression to find. 返回值:若str2是str1的子串,则返回str2在str1的首次出现的地址:如果str2

1 题目: Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Update (2014-11-02):The signature of the function had been updated to return the index instead of the pointer. If you

Brute Force算法,时间复杂度 O(mn) def strStr(haystack, needle): m = len(haystack) n = len(needle) if n == 0: return 0 if m < n: return -1 for i in range(m - n - 1): for j in range(n): if haystack[i + j] != needle[j]: break elif j == n - 1: return i return -1

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. 思路: 注意,在for循环中条件有计算得到的负数时, 一定要把计算括起来转换为int, 否则会默认转换为uchar 负数就会被误认为是一个很大的数字. ; i < - ); ++i) 实现很常规: int strStr(string haystac

Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. I wrote that function before in practice without knowing anything about strStr(), so I have a pretty clear brute-force sol

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Update (2014-11-02): The signature of the function had been updated to return the index instead of the pointer. If you still

#-*- coding: UTF-8 -*- #题意:大海捞刀,在长字符串中找出短字符串#AC源码:滑动窗口双指针的方法class Solution(object):    def strStr(self, hayStack, needle):        """        :type haystack: str        :type needle: str        :rtype: int        """        if

Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. Summary: be careful about the corner case, haystack = "", needle = "" char *strStr(char *haystack, char *ne

Implement strStr() Implement strStr(). Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack. MY: Question. 思路: 逐步查找.当出现不同时,如何回溯是关键. Solution A: class Solution { public: char *strStr(char *haystack

Learn From: http://blog.csdn.net/morley_wang/article/details/7859922 strstr(string,search) strstr() 函数搜索一个字符串在另一个字符串中的第一次出现. 该函数返回字符串的其余部分(从匹配点).如果未找到所搜索的字符串,则返回 false. string 必需.规定被搜索的字符串. search 必需.规定所搜索的字符串.如果该参数是数字,则搜索匹配数字 ASCII 值的字符. Example 1:

char * __cdecl strstr ( const char * str1, const char * str2 ) { char *cp = (char *) str1; char *s1, *s2; if ( !*str2 ) return((char *)str1); while (*cp) { s1 = cp; s2 = (char *) str2; while ( *s1 && *s2 && !(*s1-*s2) ) s1++, s2++; if (!*s

头文件:#include <string.h> strstr()函数用来检索子串在字符串中首次出现的位置,其原型为:    char *strstr( char *str, char * substr ); [参数说明]str为要检索的字符串,substr为要检索的子串. [返回值]返回字符串str中第一次出现子串substr的地址:如果没有检索到子串,则返回NULL. [函数示例]strstr()函数的使用. 复制纯文本新窗口   #include<stdio.h> #inclu