java indexof(String str) 方法复杂性
java indexof(String str)方法的复杂性是什么?我的意思是有像KMP这样的字符串匹配算法,它以线性时间运行。我正在实现一个需要在非常大的字符串中搜索大子字符串的系统,所以我可以使用java indexof(String str)方法,或者我应该实现KMP。
java indexof(String str)方法的复杂性是什么?我的意思是有像KMP这样的字符串匹配算法,它以线性时间运行。我正在实现一个需要在非常大的字符串中搜索大子字符串的系统,所以我可以使用java indexof(String str)方法,或者我应该实现KMP。
java函数复杂性是n是文本的长度,m是模式
的长度,这里是原始代码indexOf
O(n*m)
indexOf
/**
* Returns the index within this string of the first occurrence of the
* specified substring. The integer returned is the smallest value
* <i>k</i> such that:
* <blockquote><pre>
* this.startsWith(str, <i>k</i>)
* </pre></blockquote>
* is <code>true</code>.
*
* @param str any string.
* @return if the string argument occurs as a substring within this
* object, then the index of the first character of the first
* such substring is returned; if it does not occur as a
* substring, <code>-1</code> is returned.
*/
public int indexOf(String str) {
return indexOf(str, 0);
}
/**
* Returns the index within this string of the first occurrence of the
* specified substring, starting at the specified index. The integer
* returned is the smallest value <tt>k</tt> for which:
* <blockquote><pre>
* k >= Math.min(fromIndex, this.length()) && this.startsWith(str, k)
* </pre></blockquote>
* If no such value of <i>k</i> exists, then -1 is returned.
*
* @param str the substring for which to search.
* @param fromIndex the index from which to start the search.
* @return the index within this string of the first occurrence of the
* specified substring, starting at the specified index.
*/
public int indexOf(String str, int fromIndex) {
return indexOf(value, offset, count,
str.value, str.offset, str.count, fromIndex);
}
/**
* Code shared by String and StringBuffer to do searches. The
* source is the character array being searched, and the target
* is the string being searched for.
*
* @param source the characters being searched.
* @param sourceOffset offset of the source string.
* @param sourceCount count of the source string.
* @param target the characters being searched for.
* @param targetOffset offset of the target string.
* @param targetCount count of the target string.
* @param fromIndex the index to begin searching from.
*/
static int indexOf(char[] source, int sourceOffset, int sourceCount,
char[] target, int targetOffset, int targetCount,
int fromIndex) {
if (fromIndex >= sourceCount) {
return (targetCount == 0 ? sourceCount : -1);
}
if (fromIndex < 0) {
fromIndex = 0;
}
if (targetCount == 0) {
return fromIndex;
}
char first = target[targetOffset];
int max = sourceOffset + (sourceCount - targetCount);
for (int i = sourceOffset + fromIndex; i <= max; i++) {
/* Look for first character. */
if (source[i] != first) {
while (++i <= max && source[i] != first);
}
/* Found first character, now look at the rest of v2 */
if (i <= max) {
int j = i + 1;
int end = j + targetCount - 1;
for (int k = targetOffset + 1; j < end && source[j] ==
target[k]; j++, k++);
if (j == end) {
/* Found whole string. */
return i - sourceOffset;
}
}
}
return -1;
}
你可以简单地实现算法,而不使用像这样的KMP
indexOf
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Scanner;
public class Main{
int failure[];
int i,j;
BufferedReader in=new BufferedReader(new InputStreamReader(System.in));
PrintWriter out=new PrintWriter(System.out);
String pat="",str="";
public Main(){
try{
int patLength=Integer.parseInt(in.readLine());
pat=in.readLine();
str=in.readLine();
fillFailure(pat,patLength);
match(str,pat,str.length(),patLength);
out.println();
failure=null;}catch(Exception e){}
out.flush();
}
public void fillFailure(String pat,int patLen){
failure=new int[patLen];
failure[0]=-1;
for(i=1;i<patLen;i++){
j=failure[i-1];
while(j>=0&&pat.charAt(j+1)!=pat.charAt(i))
j=failure[j];
if(pat.charAt(j+1)==pat.charAt(i))
failure[i]=j+1;
else
failure[i]=-1;
}
}
public void match(String str,String pat,int strLen,int patLen){
i=0;
j=0;
while(i<strLen){
if(str.charAt(i)==pat.charAt(j)){
i++;
j++;
if(j==patLen){
out.println(i-j);
j=failure[j-1]+1;
}
} else if (j==0){
i++;
}else{
j=failure[j-1]+1;
}
}
}
public static void main(String[] args) {
new Main();
}
}