给定一个字符串数组arr []作为输入，任务是打印单词中的单词，这些单词按单词中出现的不同字符的数量排序，然后是单词的长度。
笔记：
- 如果两个单词具有相同数量的不同字符，则总字符更多的单词排在第一位。
- 如果两个单词具有相同数量的不同字符且长度相同，则必须首先打印句子中较早出现的单词。
例子：
Input: arr[] = {“Bananas”, “do”, “not”, “grow”, “in”, “Mississippi”}
Output: do in not Mississippi Bananas grow
Explanation:
After sorting by the number of unique characters and the length the output will be, do in not Mississippi Bananas grow.
Input: arr[] = {“thank”, “you”, “geeks”, “world”}
Output: you geeks thank world
Explanation:
After sorting by the number of unique characters and the length the output will be, you geeks thank world.
方法：这个想法是用 排序。
- 初始化地图数据结构，以计算给定数组的每个字符串中所有可能的不同字符。
- 然后通过传递比较器函数对数组进行排序，其中排序是通过单词中唯一字符的数量和单词长度进行的。
- 排序完成后，打印数组的字符串。
For example s = “Bananas do not grow in Mississippi”
Word Number of unique character Length of Word
do 2 2
in 2 2
not 3 3
Bananas 4 7
grow 4 4
Mississippi 4 11
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to return no of
// unique character in a word
int countDistinct(string s)
{
// Initialize map
unordered_map m;
for (int i = 0; i < s.length(); i++) {
// Count distinct characters
m[s[i]]++;
}
return m.size();
}
// Function to perform sorting
bool compare(string& s1, string& s2)
{
if (countDistinct(s1) == countDistinct(s2)) {
// Check if size of string 1
// is same as string 2 then
// return false because s1 should
// not be placed before s2
if (s1.size() == s2.size()) {
return false;
}
return s1.size() > s2.size();
}
return countDistinct(s1) < countDistinct(s2);
}
// Function to print the sorted array of string
void printArraystring(string str[], int n)
{
for (int i = 0; i < n; i++)
cout << str[i] << " ";
}
// Driver Code
int main()
{
string arr[] = { "Bananas", "do",
"not", "grow", "in",
"Mississippi" };
int n = sizeof(arr)
/ sizeof(arr[0]);
// Function call
sort(arr, arr + n, compare);
// Print result
printArraystring(arr, n);
return 0;
} Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to return no of
// unique character in a word
static int countDistinct(String s)
{
// Initialize map
Map m = new HashMap<>();
for(int i = 0; i < s.length(); i++)
{
// Count distinct characters
if (m.containsKey(s.charAt(i)))
{
m.put(s.charAt(i),
m.get(s.charAt(i)) + 1);
}
else
{
m.put(s.charAt(i), 1);
}
}
return m.size();
}
// Function to print the sorted
// array of string
static void printArraystring(String[] str,
int n)
{
for(int i = 0; i < n; i++)
{
System.out.print(str[i] + " ");
}
}
// Driver code
public static void main(String[] args)
{
String[] arr = { "Bananas", "do",
"not", "grow",
"in", "Mississippi" };
int n = arr.length;
// Function call
Arrays.sort(arr, new Comparator()
{
public int compare(String a, String b)
{
if (countDistinct(a) ==
countDistinct(b))
{
// Check if size of string 1
// is same as string 2 then
// return false because s1 should
// not be placed before s2
return (b.length() - a.length());
}
else
{
return (countDistinct(a) -
countDistinct(b));
}
}
});
// Print result
printArraystring(arr, n);
}
}
// This code is contributed by offbeat Python3
# Python3 program of the above approach
import functools
# Function to return no of
# unique character in a word
def countDistinct(s):
# Initialize dictionary
m = {}
for i in range(len(s)):
# Count distinct characters
if s[i] not in m:
m[s[i]] = 1
else:
m[s[i]] += 1
return len(m)
# Function to perform sorting
def compare(a, b):
if (countDistinct(a) == countDistinct(b)):
# Check if size of string 1
# is same as string 2 then
# return false because s1 should
# not be placed before s2
return (len(b) - len(a))
else:
return (countDistinct(a) - countDistinct(b))
# Driver Code
arr = [ "Bananas", "do", "not",
"grow", "in","Mississippi" ]
n = len(arr)
# Print result
print(*sorted(
arr, key = functools.cmp_to_key(compare)), sep = ' ')
# This code is contributed by avanitrachhadiya2155C#
// C# program of the above approach
using System;
using System.Collections;
using System.Collections.Generic;
class GFG{
// Function to return no of
// unique character in a word
static int countDistinct(string s)
{
// Initialize map
Dictionary m = new Dictionary();
for(int i = 0; i < s.Length; i++)
{
// Count distinct characters
if (m.ContainsKey(s[i]))
{
m[s[i]]++;
}
else
{
m[s[i]] = 1;
}
}
return m.Count;
}
static int compare(string s1, string s2)
{
if (countDistinct(s1) == countDistinct(s2))
{
// Check if size of string 1
// is same as string 2 then
// return false because s1 should
// not be placed before s2
return s2.Length - s1.Length;
}
else
{
return (countDistinct(s1) -
countDistinct(s2));
}
}
// Function to print the sorted array of string
static void printArraystring(string []str, int n)
{
for(int i = 0; i < n; i++)
{
Console.Write(str[i] + " ");
}
}
// Driver Code
public static void Main(string[] args)
{
string []arr = { "Bananas", "do",
"not", "grow",
"in", "Mississippi" };
int n = arr.Length;
// Function call
Array.Sort(arr, compare);
// Print result
printArraystring(arr, n);
}
}
// This code is contributed by rutvik_56 do in not Mississippi Bananas grow时间复杂度： O(n * log n)
辅助空间： O(n)