[leetcode]二分查找总结

Search for a Range

1.最简单的想法，用最普通的二分查找，找到target,然后向左右扩张，大量的重复的target，就会出现O(n)效率。

 class Solution {
    public int[] searchRange(int[] A, int target) {
        int ans[]=new int[];
      int a= bserch(A,target);
      if(a==-) {ans[]=-;ans[]=-; return ans;}
     //left
     int b=a-;
     while(b>=&&A[b]==target) b--;
     ans[]=b+;
     b=a+;

     while(b<=A.length-&&A[b]==target) b++;
     ans[]=b-;
     return ans;

    }

    public  int bserch(int A[],int target)
    {
        int low=;
        int end=A.length-;
        while(low<=end)
        {
            int mid=(low+end)>>;
            if(A[mid]==target) return mid;
            else if(A[mid]<target)  low=mid+;
            else   end=mid-;

        }

        return -;
    }

}

2.二分查找加入第一个大的位置（上届），第一个大于等于它的位置（下界）

  class Solution {
     public int[] searchRange(int[] A, int target) {
         int ans[]=new int[2];
       int a= bserch(A,target);
       if(a==-1) {ans[0]=-1;ans[1]=-1; return ans;}
      //left
      int b=a-1;
      while(b>=0&&A[b]==target) b--;
      ans[0]=b+1;
      b=a+1;

      while(b<=A.length-1&&A[b]==target) b++;
      ans[1]=b-1;
      return ans;

     }
     //其实难点就是A[mid]==target，如何调整low和high
     public  int bserch(int A[],int target)
     {
         int low=0;
         int end=A.length-1;
         while(low<=end)
         {
             int mid=(low+end)>>1;
             if(A[mid]==target) return mid;
             else if(A[mid]<target)  low=mid+1;
             else   end=mid-1;

         }

         return -1;
     }
        //第一个大于target的
     public int  upperbound(int A[],int target) // the first one larger than target
     {
         int low=0;
         int end=A.length-1;
         while(low<=end)
         {
             int mid=(low+end)>>1;
             if(A[mid]>target) end=mid-1;
             else low=mid+1;

         }
         return low;
     }
     public int  lowbound(int A[],int target) // the first one larger or equal  target
     {
         int low=0;
         int end=A.length-1;
         while(low<=end)
         {
             int mid=(low+end)>>1;
             if(A[mid]>=target) end=mid-1;
             else low=mid+1;

         }
         return low;
     }

 }