Hard 计算0到n之间2的个数 @CareerCup

一种是Brute force，O(nlogn)

另一种是找规律O(n)，见http://hawstein.com/posts/20.4.html

当某一位的数字小于2时，那么该位出现2的次数为：更高位数字x当前位数

当某一位的数字大于2时，那么该位出现2的次数为：(更高位数字+1)x当前位数

当某一位的数字等于2时，那么该位出现2的次数为：更高位数字x当前位数+低位数字+1

package Hard;
 
/**
 * Write a method to count the number of 2s between 0 and n.
 
译文：
 
写一个函数，计算0到n之间2的个数。
 *
 */
public class S18_4 {
 
	// O(n)
	public static int count2s(int n){
		int count = 0;
		int factor = 1;
		int low = 0, cur = 0, high = 0;
 
		while(n/factor != 0){
			low = n - (n/factor) * factor;		// 低位
			cur = (n/factor) % 10;				// 当前位
			high = n / (factor*10);				// 高位
 
			if(cur < 2){
				count += high * factor;
			}else if(cur > 2){
				count += (high+1) * factor;
			}else{
				count += high*factor + low + 1;
			}
 
			factor *= 10;
		}
 
		return count;
	}
 
	//=============================================
 
	public static int numberOf2s(int n) {
        int count = 0;
        while (n > 0) {
            if (n % 10 == 2) {
                    count++;
            }
            n = n / 10;
        }
        return count;
	}
 
	// Brute force way O(nlogn)
	public static int numberOf2sInRange(int n) {
        int count = 0;
        for (int i = 2; i <= n; i++) {		 // Might as well start at 2
            count += numberOf2s(i);
        }
        return count;
	}
 
	public static void main(String[] args) {
        for (int i = 0; i < 1000; i++) {
        	int b = numberOf2sInRange(i);
            int v = numberOf2sInRange(i);
            System.out.println("Between 0 and " + i + ": " + v + ", " + b);
        }
	}
 
}