package offer;
/**
 * 二分查找的思想來找到數組中重復的數字，時間復雜度在o(nlogn)-o(n^2)
 */
public class FindDuplicate3 {
 public static void main(String[] args) {
 int numbers[] = {0,1,2,3,4,4,6,7};//數組中的數 大小從0 到 numbers.length-1
 findDuplicate(numbers,0,numbers.length-1);
 }
 static void findDuplicate(int numbers[],int left,int right){
 if (numbers == null || numbers.length == 0)
 return;
 int mid; 
 while(left<=right)
 { 
 System.out.println("Find duplicate from "+left+" to "+right);
 mid=(left+right)/2;
 if(left==right)//當兩個下標值相等結束循環
 {
  if(countNumberInRange(numbers,left,right)>1)
  {  
  System.out.println(left);
  break;
  }
  else break;
 }
 //以下通過計算在指定區間數組中數字的個數與區間的長度對比來確定數組中是否有重復數字
 if(countNumberInRange(numbers,left, mid)>(mid-left+1))//如果數字區間從left到 mid的數字個數大于mid-left+1 則本區間肯定與重復數字
 {
  right=mid;
 }
 else if(countNumberInRange(numbers,mid+1, right)>(right-mid))//如果數字區間從mid+1到right的數字個數大于right-mid則本區間肯定有重復數字
 {
  left=mid+1;
 }
 else if(countNumberInRange(numbers,left, mid)==(mid-left+1) && countNumberInRange(numbers,mid+1, right)==(right-mid))
 {//因為上兩個判斷不能確定區間內是每個數字各出現了一次還是某個數字出現了兩次，所以當左右區間長度與數字個數相等時不能排除仍然有重復數字
  if(countNumberInRange(numbers,right,right)>1)//判斷最后一個數字出現次數是否是多次
  {
  System.out.println(right);
  break;
  }
  else//縮減區間
 right=right-1;
 }
 }
 
 }
 //計算數組中在from到to區間數字的個數
 static int countNumberInRange(int numbers[],int from,int to)
 {
 int count=0;
 if(numbers==null || numbers.length==0)
 return 0;
 for(int i=0;i<numbers.length;i++)
 {
 if(numbers[i]>=from && numbers[i]<=to)
 count++;
 }
 return count;
 }
}