剑指offer编程题（JAVA实现)——第31题：整数中1出现的次数

github	https://github.com/JasonZhangCauc/JZOffer

• 剑指offer编程题（JAVA实现)——第31题：整数中1出现的次数
• 求出113的整数中1出现的次数,并算出1001300的整数中1出现的次数？
• 为此他特别数了一下1~13中包含1的数字有1、10、11、12、13因此共出现6次,但是对于后面问题他就没辙了。
• ACMer希望你们帮帮他,并把问题更加普遍化,可以很快的求出任意非负整数区间中1出现的次数
• （从1 到 n 中1出现的次数）。

public class Test31 {

   public static void main(String[] args) {
   	System.out.println(NumberOf1Between1AndN_Solution(5));
   	System.out.println(NumberOf1Between1AndN_Solution(13));
   	System.out.println(NumberOf1Between1AndN_Solution(0));
   	System.out.println(NumberOf1Between1AndN_Solution(-1));
   	System.out.println(NumberOf1Between1AndN_Solution(1000));
   	System.out.println(NumberOf1Between1AndN_Solution(1));
   }

   public static int NumberOf1Between1AndN_Solution(int n) {
   	int count = 0;
   	for (int i = 0; i <= n; i++) {
   		String tmp = String.valueOf(i);
   		char[] array = tmp.toCharArray();
   		for (int j = 0; j < array.length; j++) {
   			if (array[j] == '1') {
   				count++;
   			}
   		}
   	}
   	return count;

   }
   /*
    * 此方法空间不足
    * public static int NumberOf1Between1AndN_Solution(int n) {
   	if (n < 0) {
   		return 0;
   	}
   	int count = 0;
   	int m = n;
   	if (n == 1) {
   		count++;
   		return count;
   	}
   	if (n == 0) {
   		return 0;
   	}
   	while (n >= 1) {
   		int tmp = n % 10;
   		if (tmp == 1) {
   			count++;
   		}
   		n /= 10;
   	}
   	count = count + NumberOf1Between1AndN_Solution(m - 1);
   	return count;

   }*/
}