The number of times the number that appears in the sort array: face questions fifty-three

 

Topic one: the number of times the number that appears in the sort array
method: using a binary search to find that number, and then expand the statistics about
Method Two: respectively binary search to find the first and last k K, and then subtract

 static int findk(int a[],int k) {
             if(a==null||a.length<1) return 0;
             int be=0;
             int end=a.length-1;
             int i1=-1;
             int inter;
             while(be<=end) {
                  inter=(be+end)>>1;
                  if(a[inter]<k)
                      be=inter+1;
                  else if(a[inter]>k)
                      end=inter-1;
                  else {
                      if(inter==0||a[inter-1]<k) {
                           i1=inter;
                           break;
                      }
                      else 
                           end=inter-1;               
                  }    
             }
             if(i1==-1)return 0;
             be=i1;
             end=a.length-1;
             int i2=-1;
             while(be<=end) {
                  inter=(be+end)>>1;
                  if(a[inter]<k)
                      be=inter+1;
                  else if(a[inter]>k)
                      end=inter-1;
                  else {
                      if(inter==a.length-1||a[inter+1]>k) 
                        {
                           i2=inter;
                           break;
                      }
                      else 
                           be=inter+1;                
                  }                 
             }
             if(i2>=i1)
                  return i2-i1+1;
             return 0;
         }

Title Two: the 0-n-1 deletion element
a length in ascending order in the array numbers n-1 unique numerical ranges within 0-n-1 of the n digital one and only one digit from the array, find the digital
method: direct use of binary search, if the values and subscript are equal, then the digits to the right if the value is greater than the index, and it's on a value equal to the index returns under the current standard, otherwise it on the left

 1  static int GetmissNumber(int a[]) {
 2              if(a==null||a.length<1) return 0;
 3              int be=0;
 4              int end=a.length-1;
 5              int <u>i1</u>=-1;
 6              int inter;
 7              while(be<=end) {
 8                   inter=(be+end)>>1;
 9                   if(a[inter]==inter)
10                       be=inter+1;
11                   else {
12                       if(inter==0||(a[inter]>inter&&a[inter-1]==inter-1))
13                            return inter;
14                       end=inter-1;
15                   }             
16              }        
17              return -1;
18          }

Title III: Find the same element value index
method: binary search; if the index is greater than the value that the right, if the index is less than the value at the left side, can not find -1