JavaStringTokenizer.countTokens()的时间复杂度是多少

如果您好奇的话,下面是实现:

public int countTokens() {
    int count = 0;
    int currpos = currentPosition;
    while (currpos < maxPosition) {
        currpos = skipDelimiters(currpos);
        if (currpos >= maxPosition)
            break;
        currpos = scanToken(currpos);
        count++;
    }
    return count;
}

我不太熟悉 StringTokenizer ,但假设 maxPosition skipDelimiters 和 scanToken .

OpenJDK的api说: http://www.docjar.com/html/api/java/util/StringTokenizer.java.html

public int countTokens() {
    int count = 0;
    int currpos = currentPosition;

    while (currpos < maxPosition) {    
         currpos = skipDelimiters(currpos);  
         if (currpos >= maxPosition)   
            break;    

         currpos = scanToken(currpos);    
         count++;    
     }    
     return count; 
 }

不,不是固定时间,而是 ω(n) 哪里 n

StringTokenizer的简单实现应该 O(n * d) 哪里 d 是分隔符的数目。