P**********c
发帖数: 3417 | 1 是用suffix tree还是针对这个题目有更简单直观的解法？ | w******a
发帖数: 236 | 2 Here is my code. Tested on my dev box:
const char * my_strstr(const char * s1, const char * s2)
{
int i;
for (;s1;s1++) {
for(i=0;s1&s;s1++,s2++,i++) {
if (*s1 != *s2) {
s1 -= i;
s2 -= i;
break;
}
else if (!(*(s2+1))) {
return s1-i;
}
}
}
return 0;
} | f*******t
发帖数: 7549 | | p*****u
发帖数: 310 | 4 Did not check if point is null.
【在 w******a 的大作中提到】
: Here is my code. Tested on my dev box:
: const char * my_strstr(const char * s1, const char * s2)
: {
: int i;
: for (;s1;s1++) {
: for(i=0;s1&s;s1++,s2++,i++) {
: if (*s1 != *s2) {
: s1 -= i;
: s2 -= i;
: break;
| k*j
发帖数: 153 | 5 2 solutions
1.brute force, complexity O(m*n), m is the length of the pattern, n is the
length of the string
2. KMP O(m+n) | s*****y
发帖数: 897 | 6 Should we write the kmp during the interview? Or just write the brute force? |
|
