acm专题---KMP模板

KMP的子串长n,模式串长m,复杂度o(m+n)，朴素做法的复杂度o((n-m+1)*m)

 

觉得大话数据结果上面这个讲得特别好

 

 

改进版本的KMP

leetcode

28. Implement strStr()

  

class Solution {
public:
    
    
    void getnext(string str,int next[])
{
    int i=1;
    int j=0;
    next[1]=0;
    int str0=str.length()-1;
    while(i<=str0)
    {
        if(j==0||str[i]==str[j])
        {
            ++j;
            ++i;
            next[i]=j;
        }
        else
        {
            j=next[j];
        }
    }
    
}
int kmp(string s,string t)
{
    int cnt=0;
    //int *next=new int[1000002];
    int next[1000002];
    getnext(t,next);
    int i=1;
    int j=1;
    int s0=s.length()-1,t0=t.length()-1;
    int idx=-1;
    while(i<=s0&&j<=t0)
    {
        if(j==0||s[i]==t[j])
        {
            ++i;
            ++j;
        }
        else
        {
            j=next[j];
        }
        if(j>t0)
        {
            cnt++;
            idx=i-t0-1;
            break;
            
        }
    }
    return idx;
}

int strStr(string haystack, string needle) {
    if(haystack.length()==0&&needle.length()<=haystack.length()||needle.length()==0) return 0;
    string tmpt="#",tmps="#";
    tmpt+=needle;
    tmps+=haystack;
    return kmp(tmps, tmpt);
        
}
};

 

 

class Solution {
public:
    
    void getnext(string str,int next[])
{
    int i=1;
    int j=0;
    next[1]=0;
    int str0=str.length()-1;
    while(i<=str0)
    {
        if(j==0||str[i]==str[j])
        {
            ++j;
            ++i;
            if(str[i]!=str[j])
                next[i]=j;
            else
                next[i]=next[j];
        }
        else
        {
            j=next[j];
        }
    }
    
}
int kmp(string s,string t)
{
    int cnt=0;
    //int *next=new int[1000002];
    int next[1000002];
    getnext(t,next);
    int i=1;
    int j=1;
    int s0=s.length()-1,t0=t.length()-1;
    int idx=-1;
    while(i<=s0&&j<=t0)
    {
        if(j==0||s[i]==t[j])
        {
            ++i;
            ++j;
        }
        else
        {
            j=next[j];
        }
        if(j>t0)
        {
            cnt++;
            idx=i-t0-1;
            break;
            
        }
    }
    return idx;
}


int strStr(string haystack, string needle) {
    if(haystack.length()==0&&needle.length()<=haystack.length()||needle.length()==0) return 0;
    string tmpt="#",tmps="#";
    tmpt+=needle;
    tmps+=haystack;
    return kmp(tmps, tmpt);
        
}
};

　　

　　

转载于:https://www.cnblogs.com/wuxiangli/p/5625040.html