黑马程序员__计算N的阶乘的0的个数

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public class ZeroNumelOfFactorial{/***计算N的阶乘的0的个数*//*思路一：先求出N的阶乘，然后再统计0元素的个数。*///计算N的阶乘____方法一：递归实现public static int f(int n){if(n==0) return 1;// basisreturn n*f(n-1); // recursive part}//就算N的阶乘____方法二：迭代实现/*public static int f(int n){int f = 1;for (int i = 2; i<=n ;i++)f*=i;return f;}*///统计某个指定字符在字符串中出现的次数____方法一：public static int count(String str ,String  ch){int num = 0;String max = str.length()>ch.length()?str:ch;String min = max.equals(str)?ch:str;if (max.contains(min)){num = max.length() - max.replaceAll(ch,"").length();}    return num ;}//统计某个指定字符在字符串中出现的次数____方法二：/*public static int count(String str ,String  ch){int num = 0;String max = str:ch?str.length()>ch.length;String min = ch:str?max.equals(str);char[] chs = max.toCharArray();for(ch:chs){   if(min == ch])   num++;   }return num ;}*///测试方法public static void main(String args[]){System.out.println("10!="+f(10)+"\n包含"+count(f(10)+"","0")+"个0元素");}}

public class ZeroNumelOfFactorial{/***计算N的阶乘的0的个数*//*思路一：先求出N的阶乘，然后再统计0元素的个数。*///计算N的阶乘____方法一：递归实现public static int f(int n){if(n==0) return 1;// basisreturn n*f(n-1); // recursive part}//就算N的阶乘____方法二：迭代实现/*public static int f(int n){int f = 1;for (int i = 2; i<=n ;i++)f*=i;return f;}*///统计某个指定字符在字符串中出现的次数____方法一：public static int count(String str ,String  ch){int num = 0;String max = str.length()>ch.length()?str:ch;String min = max.equals(str)?ch:str;if (max.contains(min)){num = max.length() - max.replaceAll(ch,"").length();}    return num ;}//统计某个指定字符在字符串中出现的次数____方法二：/*public static int count(String str ,String  ch){int num = 0;String max = str:ch?str.length()>ch.length;String min = ch:str?max.equals(str);char[] chs = max.toCharArray();for(ch:chs){   if(min == ch])   num++;   }return num ;}*///测试方法public static void main(String args[]){System.out.println("10!="+f(10)+"\n包含"+count(f(10)+"","0")+"个0元素");}

//运行结果

        /*

F:\JavaTest\Api\Test\007>java ZeroNumelOfFactorial10!=3628800包含2个0元素

        */

  }

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