leetcode 28 Find the Index of the First Occurrence in a String

 直接用kmp算法

class Solution {
public:
    int strStr(string haystack, string needle) {
        return kmp(haystack,needle);
    }

    int kmp(std::string &text,std::string &pattern){
        int n = text.size();
        int m = pattern.size();
        if(m == 0)
            return 0;
        std::vector<int> next;
        next.reserve(m);
        getNext(pattern,m,next);
        int j = -1;
        for(int i = 0;i < n;i++){
            while(j != -1 && text[i] != pattern[j+1]){
                j = next[j];
            }
            if(text[i] == pattern[j+1]){
                j++;
            }
            if(j == m-1)
                return i - j;
        }
        return -1;
    }

    void getNext(std::string &pattern,int len,std::vector<int> &next){
        next[0] = -1;
        int j = -1;
        for(int i = 1;i < len;i++){
            while(j != -1 && pattern[i] != pattern[j+1]){
                j = next[j];
            }
            if(pattern[i] == pattern[j+1])
                j++;
            next[i] = j;
        }
    }
};

或者另外一种写法，用的是部分匹配值(Partial Match)表

class Solution {
public:
    int strStr(string haystack, string needle) {
        return KMP(haystack,needle);
    }
    void getPM(std::string &pattern,int len,std::vector<int> &pm){
        pm[0] = 0;
        int j = 0;// j表示前缀的末尾元素的索引位置,同时它也是最长公共前后缀的长度
        //i表示后缀的末尾元素的索引位置
        for(int i = 1;i < len;i++){
            while(j > 0 && pattern[i] != pattern[j])
                j = pm[j-1];
            if(pattern[j] == pattern[i]) j++;
            pm[i] = j;
        }
    }
    int KMP(std::string &text,std::string& pattern){
        int m = pattern.size();
        if(m == 0) return 0;
        std::vector<int> pm;
        pm.resize(m);
        getPM(pattern,m,pm);
        int n = text.size();
        int j = 0;
        for(int i = 0;i < n;i++){
            while(j > 0 && text[i] != pattern[j])
                j = pm[j-1];
            if(text[i] == pattern[j]) j++;
            if(j == m)
                return i-j+1;
        }
        return -1;
    }
};

c语言实现：

int kmp(char* text_str,char* pattern_str);
int strStr(char* haystack, char* needle) {
    return kmp(haystack,needle);
}

//计算部分匹配值(Partial Match)表
void get_pm(char* pattern_str,int pm[]){
    int len = strlen(pattern_str);
    pm[0] = 0;
    int j = 0;//j表示前缀的末尾元素的位置索引，同时j也是当前最长公共前后缀的长度
    //i表示后缀的末尾元素的位置索引
    for(int i = 1;i <len;i++){
        while(j > 0 && pattern_str[i] != pattern_str[j])
            j = pm[j-1];
        if(pattern_str[i] == pattern_str[j])
            j++;
        pm[i] = j;
    }
}

int kmp(char* text_str,char* pattern_str){
    if(!text_str || !pattern_str) return -1;
    int m = strlen(pattern_str);
    if(m == 0) return 0;//空模式串约定返回0
    int *pm = (int*) malloc(m*sizeof(int));
    get_pm(pattern_str,pm);
    int n = strlen(text_str);
    int j = 0;//模式串的当前匹配位置
    for(int i = 0;i < n;i++){
        while(j > 0 && text_str[i] != pattern_str[j])
            j = pm[j-1];
        if(text_str[i] == pattern_str[j])
            j++;
        if(j == m){
            free(pm);
            return i - m + 1;
        }
    }
    free(pm);
    return -1;
}