思路1:該解法是最直觀的解法，可以先使用二分查找先找到這個元素，然後分別向左和向右遍歷，把左右相同的元素的個數都計算出來。
思路2:使用二分查找的拓展，當查找的元素有重復的時，找到元素的第一個和最後一個，這樣將可以計算出該元素有多少個重復的了。
1 #include <stdio.h>
2 #include "stdafx.h"
3
4 int GetFirstK(int* data, int length, int k, int start, int end);
5 int GetLastK(int* data, int length, int k, int start, int end);
6
7 int GetNumberOfK(int* data, int length, int k)
8 {
9 int number = 0;
10
11 if(data != NULL && length > 0)
12 {
13 int first = GetFirstK(data, length, k, 0, length - 1);
14 int last = GetLastK(data, length, k, 0, length - 1);
15
16 if(first > - 1 && last > -1)
17 number = last - first + 1;
18 }
19
20 return number;
21 }
22
23 int GetFirstK(int* data, int length, int k, int start, int end)
24 {
25 if(start > end)
26 return -1;
27
28 int middleIndex = (start + end) / 2;
29 int middleData = data[middleIndex];
30
31 if(middleData == k)
32 {
33 if((middleIndex > 0 && data[middleIndex - 1] != k) || middleIndex == 0)
34 return middleIndex;
35 else
36 end = middleIndex -1;
37 }
38 else if(middleData > k)
39 end = middleIndex - 1;
40 else
41 start = middleIndex + 1;
42
43 return GetFirstK(data, length, k, start , end);
44
45 }
46
47 int GetLastK(int* data, int length, int k, int start, int end)
48 {
49 if(start > end)
50 return -1;
51
52 int middleIndex = (start + end) /2 ;
53 int middleData = data[middleIndex];
54
55 if(middleData == k)
56 {
57 if((middleIndex < length - 1 && data[middleIndex + 1] != k ) || middleIndex == length - 1)
58 return middleIndex;
59 else
60 start = middleIndex + 1;
61 }
62 else if(middleData < k)
63 start = middleIndex + 1;
64 else
65 end = middleIndex - 1;
66
67 return GetLastK(data, length, k, start, end);
68 }
69
70 int main()
71 {
72 int data[] = {1,2,3,3,3,3,4,5};
73 int length = sizeof(data) / sizeof(int);
74 int k = 3;
75 for(int i = 0; i < length; ++i)
76 printf("%d\t", data[i]);
77 printf("\n");
78 int result = GetNumberOfK(data, length, k);
79 printf("%d出現%d次\n",k,result);
80
81 return 0;
82 }
