Implement strStr()
Implement strStr().
Returns a pointer to the first occurrence of needle in haystack, or null if needle is not part of haystack.
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, char *needle) {
int i = 0, j = 0;
while(haystack[i] != '\0' && needle[j] != '\0') {
if(haystack[i] == needle[j])
++i, ++j;
else
i = i-j+1, j = 0;
}
return needle[j] == '\0' ? haystack+(i-j) : NULL;
}
};
Solution B 与经典 KMP 算法:
对模式串 P 设置回溯数组 next. (next 只有模式串 P 的特性有关系,与目标串没有关系。)
next 的求法:
next[0] = 0; (0 位置最后匹配,下次还从此位置开始匹配(舍去从-1开始,没有意义))
next[pos] = (P[next[pos-1]] == P[pos] ? next[pos-1]+1 : 0);
(若回溯后的字符与当前字符相同,则应设置回溯位置在前回溯位置之后。否则,设置回溯位置为0)
模式串中:
当前位置 pos 不能匹配时, 回溯到 next[pos-1] 重新开始匹配。
当前位置匹配,则继续下去。
#include <iostream>
#include <vector>
using namespace std; void getNext(char *P, vector<int> &next) {
for (int i = 0; P[i] != '\0'; ++i) {
if (i == 0) next.push_back(0);
else next.push_back((P[i] == P[next[i-1]]) ? next[i-1]+1 : 0);
}
}
class Solution {
public:
char *strStr(char *haystack, char *needle) {
vector<int> next;
getNext(needle, next);
int i = 0, j = 0;
while (haystack[i] != '\0' && needle[j] != '\0') {
if (haystack[i] == needle[j]) ++i, ++j;
else if (j == 0) ++i;
else j = next[j-1];
}
return needle[j] == '\0' ? haystack+(i-j) : NULL;
}
}; int main() {
char *T = "missiissippi", *P = "issip";
cout << (Solution().strStr(T, P) ? Solution().strStr(T, P) : "NULL") << endl;
return 0;
}