Number of occurrences of a number in a sorted array

Topic Description:

Counts the number of occurrences of a number in a sorted array.

Input:

Each test case consists of two lines:

The first row has 1 integer n, which represents the size of the array. 1<=n <= 10^6.

The second row has n integers, representing the array elements, each of which is int.

The third row has 1 integer m, which indicates that there is a M-second query next. 1<=m<=10^3.

There are m lines, each with an integer k, representing the number to query.

Output:

For each test case, there is an M row output, 1 integers per line, representing the number of times the number appears in the array.

Sample input:

8

1 2 3 3 3 3 4 5

1

3

Sample output:

4

I do this problem, is to use a binary search to find the number, and then iterate over the same number, the number of statistics. The average time complexity of this practice is O (logn), the worst case is O (n), the sword refers to an offer on the train of thought is two times with a binary search to find the number of the first and last occurrence of the position, such as the time complexity of the average and the worst is O (Logn), slightly better "

Here's the code I wrote in My own mind:

#include <stdio.h> #include <stdlib.h>/* Two-point lookup method to calculate the number of key occurrences this column more highlights: http://www.bianceng.cnhttp://w  
        ww.bianceng.cn/programming/sjjg/*/int counttimesinarrays (int *arr,int len,int key) {if (Arr==null | | len<1)  
      
    return 0;  
    int start = 0;  
    int end = Len-1;  
    int mid;  
        while (start <= end) {mid = (start+end) >>1;  
        if (arr[mid] = = key) break;  
        else if (Arr[mid] > key) end = Mid-1;  
    else start = mid+1;  
    //contains 0 occurrences of int times = 0;  
        if (start <= end) {int i;  
        times = 1;  
        for (i=mid+1;i<=end;i++) if (arr[i] = = key) times++;  
    for (i=mid-1;i>=start;i--) if (arr[i] = = key) times++;  
return times;  
    int main () {int n;  while (scanf ("%d", &n)!= EOF)
    {int *arr = (int *) malloc (n*sizeof (int));  
      
        if (arr = NULL) exit (exit_failure);  
        int i;  
        for (i=0;i<n;i++) scanf ("%d", arr+i);  
        int m;  
        scanf ("%d", &m);  
            for (i=0;i<m;i++) {int k;  
            scanf ("%d", &k);  
        printf ("%d\n", Counttimesinarrays (arr,n,k));  
        Free (arr);  
    arr = NULL;  
return 0; }

Author: csdn Blog Lan pavilion Wind and Rain