(KMP) leetcode 28. Implement strStr() 字符串匹配

思路一：暴力遍历（两重循环会超时==）

class Solution {
public:
    int strStr(string haystack, string needle) {
        int h = haystack.size();
        int n = needle.size();
        if(needle.empty())
            return 0;
        for(int i=0; i< h - n +1; ++i){
            if(haystack.substr(i, n) == needle)
                return i;
        }
        return -1;
    }
};

思路：kmp算法。

参考链接：

https://www.youtube.com/watch?v=3IFxpozBs2I&t=4s

class Solution {
public:
    void prefix_table(string pattern, int prefix[], int n){
        //表长为n，pattern为要计算前后缀的字符串
        prefix[0] = 0;  
        int len = 0, i=1;  //i从索引为1处开始比较
        while(i<n){
            if(pattern[i] == pattern[len]){
                len ++;
                prefix[i] = len;
                i++;
            }
            else{
                if(len>0)
                    len = prefix[len-1];
                else{
                    prefix[i]=0;
                    i++;
                }
                    
            }  
        }
    }
    
    void move_prefix_table(int prefix[], int n){
        
        for(int i=n-1; i>0; --i)
            prefix[i] = prefix[i-1];
        prefix[0] = -1;
    }
    
    
    int strStr(string haystack, string needle) {
        int n = needle.size(), m = haystack.size();
        if(n ==0)
            return 0;
        if(m < n)
            return -1;
        int prefix[n+1];
        prefix_table(needle, prefix, n);
        move_prefix_table(prefix, n);
        int i=0, j=0; 
        //haystack[i] , m
        //needle[j] , n
        while(i<m){
            if(j == n-1 && haystack[i] == needle[j]){

            }
            if(haystack[i] == needle[j]){
                if(j == n-1)
                    return i-j;   //j = prefix[j];
                else{
                    i++;
                    j++;
                }
            }
            else{
                j = prefix[j];
                if(j==-1){
                    i++;
                    j++;
                }
            }
        }
        return -1;
    }
};