剑指offer 2.2 不修改查找重复的数字

public class Solution {
    public static void main(String[] args){
        int array[]={2,3,5,4,3,2,6,7};
        int len=8;

        duplicate(array,len);
    }
    public static int countRange(int []array, int len, int start, int end){
        if(array==null){
            return 0;
        }
        int count=0;
        for(int i=0;i<len;i++){
            if(array[i]>=start&&array[i]<=end){
                count++;
            }
        }
        return count;
    }
    public static int duplicate(int numbers[],int length) {
        if(numbers==null||length<=0){
            return -1;
        }

        int start=1;
        int end=length-1;
        while(end>=start){
            int middle=(end+start)/2;
            int count=countRange(numbers,length,start,middle);
            if(count>(middle-start+1)){
                end=middle;
            }
            else{
                start=middle+1;
            }
            if(start==end){
                if(count>1) {//如果该数字个数超过1则为重复数字
                    System.out.println(start);
                    return start;
                }
                else
                    return -1;
            }
        }
        //System.out.print(start);
        return -1;

    }
}