[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[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;

    }

    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;
    }

}

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）

 1  class Solution {
 2     public int[] searchRange(int[] A, int target) {
 3         int ans[]=new int[2];
 4       int a= bserch(A,target);
 5       if(a==-1) {ans[0]=-1;ans[1]=-1; return ans;}
 6      //left
 7      int b=a-1;
 8      while(b>=0&&A[b]==target) b--;
 9      ans[0]=b+1;
10      b=a+1;
11
12      while(b<=A.length-1&&A[b]==target) b++;
13      ans[1]=b-1;
14      return ans;
15
16
17
18
19     }
20
21     public  int bserch(int A[],int target)
22     {
23         int low=0;
24         int end=A.length-1;
25         while(low<=end)
26         {
27             int mid=(low+end)>>1;
28             if(A[mid]==target) return mid;
29             else if(A[mid]<target)  low=mid+1;
30             else   end=mid-1;
31
32         }
33
34         return -1;
35     }
36     public int  upperbound(int A[],int target) // the first one larger than target
37     {
38         int low=0;
39         int end=A.length-1;
40         while(low<=end)
41         {
42             int mid=(low+end)>>1;
43             if(A[mid]>target) end=mid-1;
44             else low=mid+1;
45
46         }
47         return low;
48     }
49     public int  lowbound(int A[],int target) // the first one larger or equal  target
50     {
51         int low=0;
52         int end=A.length-1;
53         while(low<=end)
54         {
55             int mid=(low+end)>>1;
56             if(A[mid]>=target) end=mid-1;
57             else low=mid+1;
58
59         }
60         return low;
61     }
62
63
64
65
66
67 }

[leetcode]二分查找总结