剑指offer 32. 1到n整数中2出现的次数

来源：互联网发布：域名注册机构有哪些编辑：程序博客网时间：2024/06/05 02:42

//题目：从1到n个整数中输出2出现的次数//解法1：同一每个数字中包含2的次数并相加public class Main {public static void main(String[] args) throws Exception {System.out.println(get2Count(100));}public static int get2Count(int n){int i = 1;int result = 0;while(i<=n){result = result+getNum(i);i++;}return result;}public static int getNum(int num){int result = 0;while(num != 0){if(num%10 == 2){result++;}num = num/10;}return result;}}//解法2：按照数字位数出现的次数进行计算public class Main {public static void main(String[] args) throws Exception {System.out.println(get2Count(100));}public static int get2Count(int n){String str = String.valueOf(n);return getNum(str);}public static int getNum(String str){if(str == null || str.length() == 0){return 0;}int length = str.length();int first = str.charAt(0)-'0';if(length == 1 && first>1){return 1;}if(length == 1 && first<=1){return 0;}int numFirst = 0;//第一位可能出现2的次数和两种可能if(first>2){numFirst = (int)Math.pow(10,length-1);}else if(first == 2){numFirst = Integer.parseInt(str.substring(1))+1;}int numSecond = first*(length-1)*(int)Math.pow(10,length-2);//剩余位在第一位不变的情况下可能出现2的次数，其实就是全排列int numIter = getNum(str.substring(1));//将第一位删除，迭代执行程序return numFirst+numSecond+numIter;}}

0 0