在排序数组中的出现次数 Count the number of occurrences in a sorted array

在排序数组中的出现次数 Count the number of occurrences in a sorted array

http://www.geeksforgeeks.org/count-number-of-occurrences-in-a-sorted-array/

给定一个排序数组arr[]和一个数x，求出x在arr[]中出现的次数。例子：

 Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 2  Output: 4 // x (or 2) occurs 4 times in arr[]  Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 3  Output: 1   Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 1  Output: 2   Input: arr[] = {1, 1, 2, 2, 2, 2, 3,},   x = 4  Output: -1 // 4 doesn't occur in arr[] 

解法：

1. 用二分查找法找到x在arr[]里第一次出现的位置i。
2. 用二分查找法找到x在arr[]里最后一次出现的位置j。
3. 返回 j-i+1

/* if x is present in arr[] then returns the count of occurrences of x,    otherwise returns -1. */int count(int arr[], int x, int n){  int i; // index of first occurrence of x in arr[0..n-1]  int j; // index of last occurrence of x in arr[0..n-1]       /* get the index of first occurrence of x */  i = first(arr, 0, n-1, x, n);   /* If x doesn't exist in arr[] then return -1 */  if(i == -1)    return i;      /* Else get the index of last occurrence of x. Note that we       are only looking in the subarray after first occurrence */    j = last(arr, i, n-1, x, n);           /* return count */  return j-i+1;} /* if x is present in arr[] then returns the index of FIRST occurrence    of x in arr[0..n-1], otherwise returns -1 */int first(int arr[], int low, int high, int x, int n){  if(high >= low)  {    int mid = (low + high)/2;  /*low + (high - low)/2;*/    if( ( mid == 0 || x > arr[mid-1]) && arr[mid] == x)      return mid;    else if(x > arr[mid])      return first(arr, (mid + 1), high, x, n);    else      return first(arr, low, (mid -1), x, n);  }  return -1;}  /* if x is present in arr[] then returns the index of LAST occurrence    of x in arr[0..n-1], otherwise returns -1 */int last(int arr[], int low, int high, int x, int n){  if(high >= low)  {    int mid = (low + high)/2;  /*low + (high - low)/2;*/    if( ( mid == n-1 || x < arr[mid+1]) && arr[mid] == x )      return mid;    else if(x < arr[mid])      return last(arr, low, (mid -1), x, n);    else      return last(arr, (mid + 1), high, x, n);        }  return -1;} /* driver program to test above functions */int main(){  int arr[] = {1, 2, 2, 3, 3, 3, 3};  int x =  3;  // Element to be counted in arr[]  int n = sizeof(arr)/sizeof(arr[0]);  int c = count(arr, x, n);  printf(" %d occurs %d times ", x, c);  getchar();  return 0;}