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){
            digitsList.add(n%10);
            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:

效率很低,n2.

看讨论区里的posts,发现主要问题在于对于n和n+1,我的方法是重新计算n+1是否good,浪费。如果使用dp,可以利用n是否good这个结论,从而降低复杂度。 讨论区里面还有一个lgn的dfs解法。

时间有限,留作以后优化。

 

心得2,写任何解法前,计算其时间复杂度。

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