【原创】由一道面试题说起

一个字符串参数（value）由字母（a-z,A-Z）组成，且最大字符位数为40，要求写一个函数（maxLength）返回该参数中连续相同字母的最大个数及该字母，如果最大位数有多个，则返回第一个。例：字符串“aaaddxxxxddddxxxx”，返回值为：“x,4”。
要求：请考虑代码执行的效率并注意编码的风格

package com.fish.codesample;

public class TimeTracker {
	
	private long startTime;
	public TimeTracker() {
		startTime = System.nanoTime();
	}

	public void reSet(){
		startTime = System.nanoTime();
	}
	
	public void printTime(){
		long consumingTime = System.nanoTime() - startTime;
		System.out.println("Total Time consuming : " + consumingTime);
	}
}

package com.fish.codesample;

public class FindMaxCharsSample{
    public static void main(String[] args){
    	TimeTracker tt = new TimeTracker();
    	String content = "aaaddxxxxddddxxxx";
    	findMaxLengthCharsOne(content);
    	tt.printTime();
    	tt.reSet();
    	findMaxLengthCharsTwo(content);
        tt.printTime();
        
    }
    
    private static void findMaxLengthCharsOne(String content) {
    	int max = 0;
        String result = null;

        String after = null;
        int count = 0;
        while(content.length() != 0){
            after = content.replaceAll("^(//w)//1*","");
            count = content.length() - after.length();
            if(count > max){
                max = count;
                result = content.substring(0,1);
            }
            content = after;
        }

        System.out.println(result + " , " + max);
    }
    
    private static void findMaxLengthCharsTwo(String content) {
    	char[] contentChars = content.toCharArray();
    	int length = contentChars.length;
    	int max = 0;
    	int posIndex = 0;
    	int temp = 1;
    	for(int i=1; i<length i="" if="" contentchars="" temp="" else=""> max){
    			max = temp;
    			posIndex = i;
    		}
    	}
    	
    	System.out.println(contentChars[posIndex] + " , " + max);
    }
}
</length>

Output Result as following:
x , 4
Total Time consuming : 1313574
x , 4
Total Time consuming : 54197

效率显然是第二种方法更高一些