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 still see your function signature returns a char * or String, please click the reload button to reset your code definition.
分析
这是一道模式匹配算法。给定两个字符串haystack与needle,给出needle在haystack全然匹配的首位置坐标(从0開始)。
这道题在数据结构中有解说,除了開始的简单模式匹配算法BF算法,还给出了改进后的KMP经典算法。以下分别用这两个算法实现。
AC代码
class Solution {
public:
//简单模式匹配算法(BF算法)
int strStr(string haystack, string needle) {
int len = strlen(haystack.c_str()), nl = strlen(needle.c_str());
int i = 0, j = 0;
while (i < len && j < nl)
{
if (haystack[i] == needle[j])
{
i++;
j++;
}
else{
i = i - j + 1;
j = 0;
}//else
}//while
if (j == nl)
return i - nl;
else
return -1;
}
};
KMP算法实现
class Solution {
public:
//简单模式匹配算法(BF算法)
int strStr(string haystack, string needle) {
int len = strlen(haystack.c_str()), nl = strlen(needle.c_str());
int i = 0, j = 0;
while (i < len && j < nl)
{
if (haystack[i] == needle[j])
{
i++;
j++;
}
else{
i = i - j + 1;
j = 0;
}//else
}//while
if (j == nl)
return i - nl;
else
return -1;
}
//从字符串haystack的第pos个位置開始匹配
int KMP(const string &haystack, const string &needle, int pos)
{
int len = strlen(haystack.c_str()), nl = strlen(needle.c_str());
int i = pos, j = 0;
int *n = Next(needle);
while (i < len && j < nl)
{
if (j == -1 || haystack[i] == needle[j])
{
i++;
j++;
}
else{
j = n[j];
}//else
}//while
if (j == nl)
return i - nl;
else
return -1;
}
int* Next(const string &s)
{
int i = 0, j = -1;
int next[500] ;
int len = strlen(s.c_str());
next[0] = -1;;
while (i < len)
{
while (j >-1 && s[i] != s[j])
j = next[j];
i++;
j++;
if (s[i] == s[j])
next[i] = next[j];
else
next[i] = j;
}//while
return next;
}
};