给定一个字符串数组 arr[]作为输入，任务是打印按单词中出现的不同字符的数量排序的单词，然后是单词的长度。
笔记：
- 如果两个单词的不同字符数相同，则总字符数多的单词先出现。
- 如果两个单词具有相同数量的不同字符和相同的长度，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：想法是使用 排序。
- 初始化映射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的长度来完成的。
- 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)
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