LC 660. Remove 9 【lock, hard】

Start from integer 1, remove any integer that contains 9 such as 9, 19, 29...

So now, you will have a new integer sequence: 1, 2, 3, 4, 5, 6, 7, 8, 10, 11, ...

Given a positive integer n, you need to return the n-th integer after removing. Note that 1 will be the first integer.

Example 1:

Input: 9
Output: 10

 

Hint: n will not exceed 9 x 10^8.

 

找规律题。借鉴网上的思路。

1,2,3,4,5,6,7,8,10,。。。,18,20,。。。28,30

把9拿掉以后,可以把它看成一个九进制的序列。也就是说,9进制中的数字n如果把它当作10进制来看,就是在这个序列中的第n个。

class Solution {
public:
    int newInteger(int n) {
        int base = 1;
        int ans = 0;
        while(n != 0){
            ans = ans + (n % 9) * base;
            n /= 9;
            base *= 10;
        }
        return ans;
    }
};

 

posted @ 2018-12-19 22:16  yuxihong  阅读(99)  评论(0编辑  收藏  举报