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);
}
}

}