【LeetCode】788. Rotated Digits

青旅半醒 2022-05-18 09:27 91阅读 0赞

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.

> 题目大意：给定一个数，翻转其每一位数,仍然可以成为一个有效的数字，而且不与原来的数重复，我们认为这个数是一个好数，求给定区间内好数的数目。

**分析： 对于每一个数来说，我们假设数子为abcd,可以将数子拆分为abc,d两部分。如果abc翻转完后还是abc,d翻转完后还是d，那么abcd翻转完后与原来的数一致，当abc,d两部分只有任意一个翻转后不是有效数字，abcd就翻转后就无效，只要有一个出现翻转后有效，那么abcd就可以称为好数。因此我们可以使用*动态规划*进行解题。**

**建立状态转移方程：**

**建立dp\[n+1\]，0 为无效，1为翻转后仍为原数，2表示翻转后与原数不同 。**

**dp\[i\] : 将有dp\[i/10\] 和dp\[i%10\]共同决定。**

**代码：**

int dp[] = new int[N+1];
    		 int res = 0;
    		 for(int i = 0 ; i < N+1;i++) {
    			 if(i < 10) {
    				 if (i == 0 || i == 1 || i == 8) dp[i] = 1;
    				 else if(i == 2 || i == 5 || i == 6 || i == 9) {
    					 res++;
    					 dp[i] =2;
    				 }
    			 }else {
    				 int a = dp[i/10],b = dp[i%10];
    				 if(a == 1 && b == 1) dp[i] = 1;
    				 else if((a >1 && b >=1) ||( b >1 && a>=1)) {
    					 dp[i] = 2;
    					 res++;
    				 }
    			 }
    		 }
    		 return res;

当然还可以直接暴力循环，将每一位数字转为字符串，然后逐个验证就行。