原文:
给定字符串 str ，任务是按照字符出现频率的降序打印字符。如果两个字符的频率相同，那么就按字母顺序降序排列。 例:
输入:str = " geeks forgeeks " t3】输出:t5】e–4 s–2 k–2 g–2 r–1 o–1 f–1 t13】输入:str = " bbcc " t16】输出:t18】c–2 b–2
进场 1:
- 使用一个来存储给定字符串所有元素的频率。
- 从地图中找到最大频率元素，用它的频率打印它，然后从地图中删除它。
- 当地图不为空时，重复上一步。
以下是上述方法的实现:
卡片打印处理机（card print processor 的缩写）
// c implementation of the approach
#include
using namespace std;
// function to print the characters
// of the given string in decreasing
// order of their frequencies
void printchar(string str, int len)
{
// to store the
unordered_map occ;
for (int i = 0; i < len; i )
occ[str[i]] ;
// map's size
int size = occ.size();
unordered_map::iterator it;
// while there are elements in the map
while (size--) {
// finding the maximum value
// from the map
unsigned currentmax = 0;
char arg_max;
for (it = occ.begin(); it != occ.end(); it) {
if (it->second > currentmax
|| (it->second == currentmax
&& it->first > arg_max)) {
arg_max = it->first;
currentmax = it->second;
}
}
// print the character
// alongwith its frequency
cout << arg_max << " - " << currentmax << endl;
// delete the maximum value
occ.erase(arg_max);
}
}
// driver code
int main()
{
string str = "geeksforgeeks";
int len = str.length();
printchar(str, len);
return 0;
}
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of the approach
import java.util.*;
class gfg{
// function to print the characters
// of the given string in decreasing
// order of their frequencies
static void printchar(char []arr, int len)
{
// to store the
hashmap occ = new hashmap();
for (int i = 0; i < len; i )
if(occ.containskey(arr[i]))
{
occ.put(arr[i], occ.get(arr[i]) 1);
}
else
{
occ.put(arr[i], 1);
}
// map's size
int size = occ.size();
// while there are elements in the map
while (size-- > 0)
{
// finding the maximum value
// from the map
int currentmax = 0;
char arg_max = 0;
for (map.entry it : occ.entryset())
{
if (it.getvalue() > currentmax ||
(it.getvalue() == currentmax &&
it.getkey() > arg_max))
{
arg_max = it.getkey();
currentmax = it.getvalue();
}
}
// print the character
// alongwith its frequency
system.out.print(arg_max " - "
currentmax "\n");
// delete the maximum value
occ.remove(arg_max);
}
}
// driver code
public static void main(string[] args)
{
string str = "geeksforgeeks";
int len = str.length();
printchar(str.tochararray(), len);
}
}
// this code is contributed by gauravrajput1
c
// c# implementation of the approach
using system;
using system.collections.generic;
class gfg{
// function to print the characters
// of the given string in decreasing
// order of their frequencies
static void printchar(char []arr, int len)
{
// to store the
dictionary occ = new dictionary();
for (int i = 0; i < len; i )
if(occ.containskey(arr[i]))
{
occ[arr[i]] = occ[arr[i]] 1;
}
else
{
occ.add(arr[i], 1);
}
// map's size
int size = occ.count;
// while there are elements in the map
while (size-- > 0)
{
// finding the maximum value
// from the map
int currentmax = 0;
char arg_max = (char)0;
foreach (keyvaluepair it in occ)
{
if (it.value > currentmax ||
(it.value == currentmax &&
it.key > arg_max))
{
arg_max = it.key;
currentmax = it.value;
}
}
// print the character
// alongwith its frequency
console.write(arg_max " - "
currentmax "\n");
// delete the maximum value
occ.remove(arg_max);
}
}
// driver code
public static void main(string[] args)
{
string str = "geeksforgeeks";
int len = str.length;
printchar(str.tochararray(), len);
}
}
// this code is contributed by princi singh
java 描述语言
output
e - 4
s - 2
k - 2
g - 2
r - 1
o - 1
f - 1
方法 2 : 我们将制作一个数组 arr ，其大小比给定字符串长度的大小大一，我们将在其中存储频率等于 arr 索引的字符列表，并遵循以下步骤:
- 使用给定字符串中的字符数组制作频率图。
- 遍历频率数组，如果它的值大于零就说 k.
- 在 arr 的第 k 个索引上，将其字符值存储在索引 0 处的列表中(因为如果频率相同，我们需要字母的降序)。
- 遍历从后开始排列，因为如果索引处列表不是空的，我们首先需要更高的频率，而不是打印它的频率和字符。
上述方法的实施:
java 语言(一种计算机语言，尤用于创建网站)
// java implementation of above approach
import java.util.*;
class gfg {
// driver code
public static void main(string[] args)
{
string str = "geeksforgeeks";
printchar(str);
}
@suppresswarnings("unchecked")
// function to print the characters
// of the given string in decreasing
// order of their frequencies
public static void printchar(string str)
{
// initializing array of list type.
list[] arr = new list[str.length() 1];
for (int i = 0; i <= str.length(); i ) {
// initializing list of type character.
arr[i] = new arraylist<>();
}
int[] freq = new int[256];
// mapking frequency map
for (int i = 0; i < str.length(); i ) {
freq[(char)str.charat(i)] ;
}
// traversing frequency array
for (int i = 0; i < 256; i ) {
if (freq[i] > 0) {
// if frequency array is greater than zero
// then storing its character on
// i-th(frequency of that character) index
// of arr
arr[freq[i]].add(0, (char)(i));
}
}
// traversing arr from backwards as we need greater
// frequency character first
for (int i = arr.length - 1; i >= 0; i--) {
if (!arr[i].isempty()) {
for (char ch : arr[i]) {
system.out.println(ch "-" i);
}
}
}
}
}
output
e-4
s-2
k-2
g-2
r-1
o-1
f-1
时间复杂度: o(n)，n 为给定字符串的长度
辅助空间: o(n)
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