二分查找

参见编程珠玑chap4 chap9，和编程之美3.3

 

 

 

给定一个有序（不降序）数组arr,，

1. 求任意一个i使得arr[i]等于t，不存在返回-1

 

不变式：x[low]< =t<=x[high]
 

 

int biSearch(int *arr, int low, int high, int t)



{



    int mid;



 



    while(low<=high)



    {



        mid=low+(high-low)/2;



        if(arr[mid]==t)



        {



            return mid;



        }



        else if(arr[mid]>t)



        {



           high=mid-1;



        }



        else



        {



            low=mid+1;



        }



    }



    return -1;



}

　　

 

2. 如存在重复元素，求最小的i使得arr[i]等于t，不存在返回-1

 

不变式：x[low]< t<=x[high]， return high
 

 1 int biSearch(int *arr, int start, int end, int t)

 2 

 3 {

 4 

 5     int high, low, mid;

 6 

 7     //x[l]<t=<x[u]

 8 

 9     high=end+1;

10 

11     low=start-1;

12 

13     while(low+1!=high)

14 

15     {

16 

17         mid = low + (high-low)/2;

18 

19         if(t<=arr[mid])

20 

21         {

22 

23             high=mid;

24 

25         }

26 

27         else

28 

29         {

30 

31             low=mid;

32 

33         }

34 

35     }

36 

37     if(high>=end+1 || arr[high]!=t)

38 

39     {

40 

41         return -1;

42 

43     }

44 

45     else

46 

47     {

48 

49         return high;

50 

51     }

52 

53 }

 

 

 

3. 如存在重复元素，求最大的i使得arr[i]等于t，不存在返回-1

 

不变式：x[low]<= t<x[high]， return low

 

 1 int biSearch(int *arr, int start, int end, int t)

 2 

 3 {

 4 

 5     int high, low, mid;

 6 

 7     //x[l]<=t<x[u]

 8 

 9     high=end+1;

10 

11     low=start-1;

12 

13     while(low+1!=high)

14 

15     {

16 

17         mid = low + (high-low)/2;

18 

19         if(t<arr[mid])

20 

21         {

22 

23             high=mid;

24 

25         }

26 

27         else

28 

29         {

30 

31             low=mid;

32 

33         }

34 

35     }

36 

37     if(low<=start-1 || arr[low]!=t)

38 

39     {

40 

41         return -1;

42 

43     }

44 

45     else

46 

47     {

48 

49         return low;

50 

51     }

52 

53 }

 

 

4. 求最大的i使得arr[i]小于t，不存在返回-1；

 

不变式： x[low] <t<=x[high], return low

 

 1 int biSearch(int *arr, int start, int end, int t)

 2 

 3 {

 4 

 5     int high, low, mid;

 6 

 7     //x[l]<t=<x[u]

 8 

 9     high=end+1;

10 

11     low=start-1;

12 

13     while(low+1!=high)

14 

15     {

16 

17         mid = low + (high-low)/2;

18 

19         if(t<=arr[mid])

20 

21         {

22 

23             high=mid;

24 

25         }

26 

27         else

28 

29         {

30 

31             low=mid;

32 

33         }

34 

35     }

36 

37     if(low<=start-1 || arr[low]>t)

38 

39     {

40 

41         return -1;

42 

43     }

44 

45     else

46 

47     {

48 

49         return low;

50 

51     }

52 

53 }

 

 

 

5. 求最小的i使得arr[i]大于t，不存在返回-1；

不变式：x[low]<=t<x[high]，return high

略

 

测试程序：

 1 #include <iostream>

 2 

 3 using namespace std;

 4 

 5 

 6 void init(int *a, int n)

 7 {

 8     for(int i=0; i<n; i++)

 9     {

10         a[i]=(i)/3+1;

11     }

12 }

13 

14 void print(int *a, int n)

15 {

16     for(int i=0; i<n; i++)

17     {

18         cout<<a[i]<<" ";

19     }

20     cout<<endl;

21 }

22 

23 int biSearch(int *arr, int start, int end, int t)

24 {

25     int high, low, mid;

26     //x[l]<t=<x[u]

27     high=end+1;

28     low=start-1;

29     while(low+1!=high)

30     {

31         mid = low + (high-low)/2;

32         if(t<=arr[mid])

33         {

34             high=mid;

35         }

36         else

37         {

38             low=mid;

39         }

40     }

41     if(low<=start-1 || arr[low]>t)

42     {

43         return -1;

44     }

45     else

46     {

47         return low;

48     }

49 }

50 

51 int main()

52 {

53     int n=5;

54     int arr[n];

55 

56     init(arr, n);

57     print(arr, n);

58     for(int i=-2; i<n+1; i++)

59         cout<<i<<":"<<biSearch(arr, 0, n-1, i)<<endl;

60 

61 

62     return 0;

63 }