原文:
给定一个数组 arr ，任务是按照降序打印数组的元素及其频率。 例:
输入: arr[] = {1，3，3，3，4，4，5} 输出: 5 发生 1 次 4 发生 2 次 3 发生 3 次 1 发生 1 次 输入: arr[] = {1，1，1，2，3，4，9，10} 输出: 10 发生 1 次 9 发生 2 次【t15
简单方法:使用一些(例如)，以降序存储元素，然后用它的计数逐个打印元素，然后从数据结构中删除它。对于所使用的数据结构，时间复杂度为 0(n 对数 n)，辅助空间为 0(n)。 以下是上述方法的实现:
卡片打印处理机（card print processor 的缩写）
// c program to print the elements in
// descending along with their frequencies
#include
using namespace std;
// function to print the elements in descending
// along with their frequencies
void printelements(int a[], int n)
{
// a multiset to store elements in decreasing order
multiset > ms;
// insert elements in the multiset
for (int i = 0; i < n; i ) {
ms.insert(a[i]);
}
// print the elements along with their frequencies
while (!ms.empty()) {
// find the maximum element
int maxel = *ms.begin();
// number of times it occurs
int times = ms.count(maxel);
cout << maxel << " occurs " << times << " times\n";
// erase the maxel
ms.erase(maxel);
}
}
// driver code
int main()
{
int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = sizeof(a) / sizeof(a[0]);
printelements(a, n);
return 0;
}
output:
10 occurs 1 times
9 occurs 2 times
4 occurs 1 times
3 occurs 1 times
2 occurs 1 times
1 occurs 3 times
高效方法: ，然后开始打印元素及其频率。 以下是上述方法的实施:
c
// c program to print the elements in
// descending along with their frequencies
#include
using namespace std;
// function to print the elements in descending
// along with their frequencies
void printelements(int a[], int n)
{
// sorts the element in decreasing order
sort(a, a n, greater());
int cnt = 1;
// traverse the array elements
for (int i = 0; i < n - 1; i ) {
// prints the number and count
if (a[i] != a[i 1]) {
cout << a[i] << " occurs " << cnt << " times\n";
cnt = 1;
}
else
cnt = 1;
}
// prints the last step
cout << a[n - 1] << " occurs " << cnt << " times\n";
}
// driver code
int main()
{
int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = sizeof(a) / sizeof(a[0]);
printelements(a, n);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java program to print the elements in
// descending along with their frequencies
import java.util.*;
class gfg
{
// function to print the elements in descending
// along with their frequencies
static void printelements(int a[], int n)
{
// sorts the element in decreasing order
arrays.sort(a);
a = reverse(a);
int cnt = 1;
// traverse the array elements
for (int i = 0; i < n - 1; i )
{
// prints the number and count
if (a[i] != a[i 1])
{
system.out.print(a[i] " occurs "
cnt " times\n");
cnt = 1;
}
else
cnt = 1;
}
// prints the last step
system.out.print(a[n - 1] " occurs "
cnt " times\n");
}
static int[] reverse(int a[])
{
int i, n = a.length, t;
for (i = 0; i < n / 2; i )
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// driver code
public static void main(string[] args)
{
int a[] = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = a.length;
printelements(a, n);
}
}
// this code is contributed by princiraj1992
python 3
# python3 program to print the elements in
# descending along with their frequencies
# function to print the elements in
# descending along with their frequencies
def printelements(a, n) :
# sorts the element in decreasing order
a.sort(reverse = true)
cnt = 1
# traverse the array elements
for i in range(n - 1) :
# prints the number and count
if (a[i] != a[i 1]) :
print(a[i], " occurs ", cnt, "times")
cnt = 1
else :
cnt = 1
# prints the last step
print(a[n - 1], "occurs", cnt, "times")
# driver code
if __name__ == "__main__" :
a = [ 1, 1, 1, 2,
3, 4, 9, 9, 10 ]
n = len(a)
printelements(a, n)
# this code is contributed by ryuga
c
// c# program to print the elements in
// descending along with their frequencies
using system;
class gfg
{
// function to print the elements in descending
// along with their frequencies
static void printelements(int []a, int n)
{
// sorts the element in decreasing order
array.sort(a);
a = reverse(a);
int cnt = 1;
// traverse the array elements
for (int i = 0; i < n - 1; i )
{
// prints the number and count
if (a[i] != a[i 1])
{
console.write(a[i] " occurs "
cnt " times\n");
cnt = 1;
}
else
cnt = 1;
}
// prints the last step
console.write(a[n - 1] " occurs "
cnt " times\n");
}
static int[] reverse(int []a)
{
int i, n = a.length, t;
for (i = 0; i < n / 2; i )
{
t = a[i];
a[i] = a[n - i - 1];
a[n - i - 1] = t;
}
return a;
}
// driver code
public static void main(string[] args)
{
int []a = { 1, 1, 1, 2, 3, 4, 9, 9, 10 };
int n = a.length;
printelements(a, n);
}
}
// this code is contributed by princiraj1992
服务器端编程语言（professional hypertext preprocessor 的缩写）
java 描述语言
output:
10 occurs 1 times
9 occurs 2 times
4 occurs 1 times
3 occurs 1 times
2 occurs 1 times
1 occurs 3 times
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