【剑指offer】数字在排序数组中出现的次数

2013-09-02 16:28:35

找出数字在排序数组中出现的次数。

注意几点：

1. 一开始试图用size_t类型表示数组的下标begin、end，到那时这样做在end = 0时，end - 1是size_t类型的最大值，仍然满足begin <= end，但此时将会对sortedArray数组中下标为size_t类型的最大值的元素，会出现访问越界；因此，对于数组小标，不要为了保证是整数二用size_t类型，用int类型比较好。
2. 若用int型表示，就不需要用STATUS的状态标志，下面的程序中没有修改这一点。

代码：

#include <iostream>
#include <cassert>
using namespace std;

typedef int DataType;

const bool SuccessToFind = true;
const bool FailToFind = false;

bool STATUS = SuccessToFind;

//找出第一个K出现的位置的下标
int GetFirstK(DataType *sortedArray,int begin,int end,const DataType data)
{
    /*assert(NULL != sortedArray);
    assert(begin <= end);*/

    STATUS = FailToFind;
    int mid = 0;

    while (begin <= end)
    {
        mid = begin + (end - begin) / 2;
        if (sortedArray[mid] == data)
        {
            if (mid == begin || sortedArray[mid - 1] != data)
            {
                STATUS = SuccessToFind;
                return mid;
            }
            else
            {
                end = mid - 1;
            }
        }
        else if (sortedArray[mid] < data)
        {
            begin = mid + 1;
        }
        else
        {
            end = mid - 1;   //mid为0时，若end为size_t类型，会出错
        }
    }

    STATUS = FailToFind;
    return 0;
}

//找出最后一个K出现的位置的下标
int GetLastK(DataType *sortedArray,int begin,int end,const DataType data)
{
    /*assert(NULL != sortedArray);
    assert(begin <= end);*/

    STATUS = FailToFind;
    int mid = 0;

    while (begin <= end)
    {
        mid = begin + (end - begin) / 2;
        if (sortedArray[mid] == data)
        {
            if (mid == end || sortedArray[mid + 1] != data)
            {
                STATUS = SuccessToFind;
                return mid;
            }
            else
            {
                begin = mid + 1;
            }
        }
        else if (sortedArray[mid] < data)
        {
            begin = mid + 1;
        }
        else
        {
            end = mid - 1;
        }
    }

    STATUS = FailToFind;
    return 0;
}

//返回K出现的次数
int GetNumberOfK(DataType *sortedArray,int len,const DataType data)
{
    assert(NULL != sortedArray);
    assert(len >= 1);

    size_t begin = GetFirstK(sortedArray,0,len - 1,data);
    size_t end = GetLastK(sortedArray,0,len - 1,data);

    if (STATUS == SuccessToFind)
    {
        return (end - begin + 1);
    }
    else
    {
        return 0;
    }
}

//测试GetNumberOfK
void TestGetNumberOfK()
{
    DataType sortedArray[10] = {1,2,3,3, 3,3,4,4, 5,6};
    size_t len = 10;
    DataType data = 0;

    cout<<"please enter the data to find ,end with ctrl+z "<<endl;
    while (cin>>data)
    {
        cout<<"the number of "<<data<<" in array is : "<<GetNumberOfK(sortedArray,len,data)<<endl;
        cout<<"please enter the data to find ,end with ctrl+z "<<endl;
    }

}

int main()
{
    TestGetNumberOfK();
    return 0;
}

测试结果：

please enter the data to find ,end with ctrl+z
3
the number of 3 in array is : 4
please enter the data to find ,end with ctrl+z
4
the number of 4 in array is : 2
please enter the data to find ,end with ctrl+z
2
the number of 2 in array is : 1
please enter the data to find ,end with ctrl+z
1
the number of 1 in array is : 1
please enter the data to find ,end with ctrl+z
-1
the number of -1 in array is : 0
please enter the data to find ,end with ctrl+z
90
the number of 90 in array is : 0
please enter the data to find ,end with ctrl+z
^Z
请按任意键继续. . .