Replace all occurrence of the given pattern to ‘X’.
For example, given that the pattern=”abc”, replace “abcdeffdfegabcabc” with “XdeffdfegX”. Note that multiple occurrences of abc’s that are contiguous will be replaced with only one ‘X’.

First, it is not clear whether the problem mentions an in-place replacement or not, so be sure to ask this question during an interview. Many interview questions asked are purposely ambiguous. It is expected that the candidate ask thought-provoking questions of the interviewer in order to better answer the question. Here, we will assume that it is an in-place replacement.

思路：
1、還是用一前一后的兩個指針，前一個指針用于遍歷，后一個指針用于修改值。
2、如果pFast當前所指的位置可以匹配，pFast向前移動Pattern長度，并且記下匹配的信息，直到找到第一個不能匹配的點。
3、如果標記顯示有子串匹配。則將pSlow替換為指定的字符。
4、如果pFast還沒有到Str的尾部，則將pFast賦給pSlow，因為pFast是剛剛找到的第一個未匹配的點。
?流程圖

bool IsMatch(const char *Str, const char *Pattern) { assert(Str && Pattern); while(*Pattern) { if(*Str++ != *Pattern++) { return false; } } return true; } void StrReplace(char *Str, const char *Pattern, const char ReplaceChar) { assert(Str && Pattern); int nLen = strlen(Pattern); if(nLen <= 0) { return; } char *pSlow, *pFast; pSlow = pFast = Str; bool isMatch = false; while(*pFast != '\0') { isMatch = false; while(IsMatch(pFast, Pattern))//略過與模式串匹配的部分，定位第一個不匹配的字符 { isMatch = true; pFast += nLen; } if(isMatch)//如果有匹配的，替換成相應字符 { *pSlow++ = ReplaceChar; } if(*pFast != '\0')//如果pFast沒有指向末尾 { *pSlow++ = *pFast++; //等同于*(pSlow++) = *(pFast++);因為++的優先級高于* } } *pSlow = '\0';//加上結束標志 }

測試代碼

#include<stdio.h> #include<assert.h> #include<string.h> int main() { const int MAX_N = 50; char Pattern[MAX_N]; char Str[MAX_N]; while(gets(Str) && gets(Pattern)) { StrReplace(Str, Pattern, 'X'); puts(Str); } return 1;

總結

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