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法律研究不改變母音和子音的相對位置就能排列單詞嗎？
假設我們有一個字串包含 n 個元素（n < 10）。我們必須找到在不改變母音和子音的相對位置的情況下排列字串的方法數。
方法很簡單。我們必須數出給定字串中母音和子音的數量，然後我們必須找到我們僅能用母音排列的方法數，隨後找到僅用子音排列的方法數，最後將這兩個結果相乘以取得總的方法數。
演算法
arrangeWayCount(str)
Begin define an array ‘freq’ to store frequency. count and place frequency of each characters in freq array. such that freq[‘0’] will hold frequency of letter ‘a’, freq[1] will hold frequency of ‘b’ and so on. v := number of vowels, and c := number of consonants in str vArrange := factorial of v for each vowel v in [a, e, i, o, u], do vArrange := vArrange / factorial of the frequency of v done cArrange := factorial of c for each consonant con, do cArrange := cArrange / factorial of the frequency of con done return vArrange * cArrange End
示例
#include <iostream>
using namespace std;
long long factorial(int n){
if(n == 0 || n == 1)
return 1;
return n*factorial(n-1);
}
long long arrangeWayCount(string str){
long long freq[27] = {0}; //fill frequency array to 0
int v = 0, c = 0;
for (int i = 0; i < str.length(); i++) {
freq[str[i] - 'a']++;
if (str[i] == 'a' || str[i] == 'e' || str[i] == 'i' || str[i] == 'o' || str[i] == 'u') {
v++;
}else
c++;
}
long long arrangeVowel;
arrangeVowel = factorial(v);
arrangeVowel /= factorial(freq[0]); // vowel a
arrangeVowel /= factorial(freq[4]); // vowel e
arrangeVowel /= factorial(freq[8]); // vowel i
arrangeVowel /= factorial(freq[14]); // vowel o
arrangeVowel /= factorial(freq[20]); // vowel u
long long arrangeConsonant;
arrangeConsonant = factorial(c);
for (int i = 0; i < 26; i++) {
if (i != 0 && i != 4 && i != 8 && i != 14 && i != 20)
arrangeConsonant /= factorial(freq[i]); //frequency of all characters except vowels
}
long long total = arrangeVowel * arrangeConsonant;
return total;
}
main() {
string str = "computer";
long long ans = arrangeWayCount(str);
cout << "Possible ways to arrange: " << ans << endl;
}輸出
Possible ways to arrange: 720
更新於： 2020 年 7 月 2 日
106 次瀏覽
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