Leetcode 788: Rotated Digits

问题描述：
x is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from x. Each digit must be rotated - we cannot choose to leave it alone.

A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other (on this case they are rotated in a different direction, in other words 2 or 5 gets mirrored); 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid.

Now given a positive number n, how many numbers x from 1 to n are good?

解释说明：我们要把一个数的每一位旋转180度，若旋转后的数字依然是数字，而且与原数字不同，则我们认为原数字是好数字。
对应关系：1->1, 2->5, 3->inValid, 4->inVaild, 5->2, 6->9, 7->inValid, 8->8, 9->6, 0->0
比如说126旋转以后是159, 且159！=126，所以126是好数字；232含有不能有效旋转的3，故不是好数字；8811旋转后还是8811，故也不是好数字

代码如下：

class Solution {public int rotatedDigits(int n) {int counter=0;for(int i=1; i<=n; i++){if(isValid(i))  counter++;}return counter;}private boolean isValid(int n){String str_original=Integer.toString(n);String str_new="";for(int i=0; i<str_original.length(); i++){if((str_original.charAt(i)=='3')||(str_original.charAt(i)=='4')||(str_original.charAt(i)=='7')){return false;}else{str_new=str_new+rotating(str_original.charAt(i));}}return !str_original.equals(str_new);}private char rotating(char a){if(a=='0')  return '0';else if(a=='1')  return '1';else if(a=='2')  return '5';else if(a=='5')  return '2';else if(a=='6')  return '9';else if(a=='8')  return '8';else return '6';}
}

时间复杂度: 因为每个数字n的长度与其数值是log(n)关系，所以时间复杂度是log1+log2+log3+log4+…+logn=log(n!)=O(nlogn)

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