Question

可能重复：
What is the cost / complexity of a String.indexof() function call

java indexof（String str）方法的复杂性是什么？我的意思是像KMP这样的字符串匹配算法在线性时间内运行。我正在实现一个需要在一个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。

Answer 1

Java indexOf O(m*n)的复杂性为n，其中m和{{1}}分别是搜索字符串和模式的长度。

您可以采取的措施是提高复杂性，例如使用implementation算法智能地跳过比较字符串中与模式不匹配的逻辑部分。

Answer 2

java indexOf函数复杂度为O(n*m)，其中n是文本的长度，m是模式的长度
这是indexOf原始代码

   /**
     * 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 &gt;= 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;
    }

您可以简单地实现KMP算法而不使用indexOf这样的

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();
    }
}

java indexof（String str）方法的复杂性

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