leetcode 二分查找

https://oj.leetcode.com/problems/search-for-a-range/就是一个二分查找，没事练练手

 public class Solution {
     public int[] searchRange(int[] A, int target) {
         int a[]=new int[2];
         int ans=bSearch(A,target);
         if(ans==-1)
         {
             a[0]=-1;
             a[1]=-1;
             return a;

         }
         else
         {
             int beg=ans;
             int end=ans;
             while(beg>=0&&(A[beg]==A[ans]))
             {
                 beg--;

             }
             while(end<A.length&&(A[end]==A[ans]))
             {
                 end++;
             }
             a[0]=beg+1;
             a[1]=end-1;
             return a;

         }

     }
     public int bSearch(int[] A,int target)
     {
         int low=0;
         int high=A.length-1;
         while(low<=high)
         {
             int mid =(low+high)/2;
             if(A[mid]==target)
             {
                 return mid;

             }
             else if(A[mid]>target)
             {

                 high=mid-1;

             }
             else
             {
                 low=mid+1;
             }

         }

         return -1;
     }

 }

2.另外一种二分查找 beg=0 end=len;  end指向的是最后一个元素的后面，

 public class Solution {
     public int[] searchRange(int[] A, int target) {
         int a[]=new int[2];
         int ans=bSearch(A,target);
         if(ans==-1)
         {
             a[0]=-1;
             a[1]=-1;
             return a;

         }
         else
         {
             int beg=ans;
             int end=ans;
             while(beg>=0&&(A[beg]==A[ans]))
             {
                 beg--;

             }
             while(end<A.length&&(A[end]==A[ans]))
             {
                 end++;
             }
             a[0]=beg+1;
             a[1]=end-1;
             return a;

         }

     }
     public int bSearch(int[] A,int target)
     {
         int low=0;
         int high=A.length;
         while(low<high)  //low<high不能等于
         {
             int mid =(low+high)/2;
             if(A[mid]==target)
             {
                 return mid;

             }
             else if(A[mid]>target)
             {

                 high=mid;

             }
             else
             {
                 low=mid+1;
             }

         }

         return -1;
     }

 }

3.第一种方法的递归方式

 public class Solution {
     public int[] searchRange(int[] A, int target) {
         int a[]=new int[2];
         int ans=bSearch(A,target,0,A.length);
         if(ans==-1)
         {
             a[0]=-1;
             a[1]=-1;
             return a;

         }
         else
         {
             int beg=ans;
             int end=ans;
             while(beg>=0&&(A[beg]==A[ans]))
             {
                 beg--;

             }
             while(end<A.length&&(A[end]==A[ans]))
             {
                 end++;
             }
             a[0]=beg+1;
             a[1]=end-1;
             return a;

         }

     }
     public int bSearch(int[] A,int target,int low,int high)
     {

         while(low<high)
         {
             int mid =(low+high)/2;
             if(A[mid]==target)
             {
                 return mid;

             }
             else if(A[mid]>target)
             {

                 bSearch(A,target,low,mid);

             }
             else
             {
                 bSearch(A,target,mid,high);
             }

         }

         return -1;
     }

 }