# LeetCode 788. Rotated Digits

question:

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; 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?

Example:
Input: 10
Output: 4
Explanation:
There are four good numbers in the range [1, 10] : 2, 5, 6, 9.
Note that 1 and 10 are not good numbers, since they remain unchanged after rotating.

Note:

N  will be in range [1, 10000].

try:

class Solution {
public int rotatedDigits(int N) {
int counts=0;
for(int i=1;i<=N;i++){
counts=valid(i)? counts+1:counts;
}
return counts;
}

public boolean valid(int n){
ArrayList<Integer> digitsList = new ArrayList<Integer>();
while(n!=0){
n=n/10;

}

boolean contains = false;
for(Integer integer:digitsList){
if(integer.equals(3)||integer.equals(4)||integer.equals(7)){
return false;
}
if(integer.equals(2)||integer.equals(5)||integer.equals(6)||integer.equals(9)){
contains=true;
}
//return false;
}
return contains;
}
}

result:

re-try:

class Solution {
public int rotatedDigits(int N) {
int counts=0;
for(int i=1;i<=N;i++){
counts=valid(i)? counts+1:counts;
}
return counts;
}

public boolean valid(int n){

String s=((Integer)n).toString();
char[] charArray = new char[s.length()];
for(int i=0;i<s.length();i++){
charArray[i]=s.charAt(i);
}

boolean contains = false;
for(char c:charArray){
if(c=='3'||c=='4'||c=='7'){
return false;
}
if(c=='2'||c=='5'||c=='6'||c=='9'){
contains=true;
}
//return false;
}
return contains;
}
}

result:

conclusion:

posted @ 2018-04-29 11:39  Zhao_Gang  阅读(107)  评论(0编辑  收藏  举报