leetcode_28_ Implement strStr() (easy)

Imple

ment strStr()

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.

解题:

要考虑“”，""情况，输出为0

最初的写法,暴力法:
class Solution {
public:
    int strStr(string haystack, string needle) {
        if(needle.length() == 0) {
            if(haystack.length() == 0)return 0;
            return -1;
        }
        for(int i =0;i<=haystack.length()-needle.length();i++){
            if(haystack.at(i)==needle.at(0)){
                int y = i;
                int equation = 0;
                for(int j = 0;j<needle.length();j++,y++){
                    if(haystack.at(y)!=needle.at(j)){
                        equation = 1;
                        break;
                    }
                }
                if(!equation){
                    return i;
                }
            }
        }
        return -1;
    }
     
};
超时
然后:

这道题的直接思路就是暴力查找，就是一个字符一个字符地进行匹配。不多说了，如果有兴趣，可以去研究这几个O(m+n)的算法：Rabin-Karp算法、KMP算法和Boyer-Moore算法。

时间复杂度：O(mn)

空间复杂度：O(1)

#include <iostream>

using namespace std;class Solution {public:    int strStr(string haystack, string needle) {        if(haystack.length()<needle.length())return -1;//没有此句，会超时        unsigned long al = haystack.length();        unsigned long bl = needle.length();        for(int i =0,j;i<=al-bl;i++){            for( j = 0;j<needle.length()&&haystack.at(i+j)==needle.at(j);j++);                if(j==bl){                    return i;                }        }        return -1;    }    };int main(int argc, const char * argv[]) {    Solution c;    string a;    string b;    while (1) {        cout<<"start"<<endl;        cin>>a;        cin>>b;        cout<<c.strStr(a, b)<<endl;    }      return 0;}