假设我们有一个包含 n 个元素的字符串 (n
方法很简单。我们必须计算给定字符串中元音和辅音的数量，然后我们必须找到仅可以排列元音的方式有多少种，然后找到仅排列辅音的方式的数量，然后将这两个结果相乘得到总路数。
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 * cArrangeEnd
示例
#include 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;}
输出
Possible ways to arrange: 720
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