LeetCode788. Rotated Digits旋转的数字

LeetCode788. Rotated Digits旋转的数字

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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].

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分析：

求1-N中不包含3，4，7，包含2，5，6，9的数字的个数。

重要测试用例：
1-10
459
1999-2010

思路：
3456
1）求      0-3000中的个数,n1
2）求3001-3400中的个数,n2
3）求3401-3450中的个数,n3
4）求3451-3456中的个数,n4
number= n1+ n2 + n3 + n4

难点：
高位为2，5，6，9和高位为0,1,8，Good的个数不同。

区分：
|N| highIsGood | !highIsGood|
|:-|:-|
|0 | 0 |1|
| 1| 0 | 2|
| 2 | 1 | 3|
| 3 | 1 | 3|
| 4 | 1 | 3|
|5 | 2 | 4|
|6 | 3 | 5|
| 7 | 3 | 5|
|8 | 3 | 6|
|9 | 4 | 7|

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代码

class Solution {
public:
	int rotatedDigits(int N) {
		// N [1, 10000], test case: 1-10, 1999-2010, 345, 467
		int sum = 0, temp = N, high, unit;
		bool highIsGood;

		highIsGood = false;
		unit = getUnit(temp);
		high = temp / unit;		

		do
		{			
			sum += getSum1(high, unit, highIsGood);
			// test case: 328
			if (3 == high || 4 == high || 7 == high)
				break;
			if(!highIsGood) highIsGood = (2 == high || 5 == high || 6 == high || 9 == high);
			temp -= high * unit;
			// test case: 200
			if (temp == 0 && unit > 1 && highIsGood) sum += 1;
			unit = getUnit(temp);
			high = temp / unit;
		}while (temp);

		return sum;
	}

	int getUnit(int n){
		int unit = 1;
		while (n / 10){
			n /= 10;
			unit *= 10;
		}
		return unit;
	}

	int getSum1(int high, int unit, bool highIsGood){
		switch (high){
		case 1: return 1 * getSum2(unit, highIsGood) + highIsGood*(unit == 1);
		case 2: return 2 * getSum2(unit, highIsGood) + (unit == 1);
		case 3:
		case 4: return 2 * getSum2(unit, highIsGood) + 1 * getSum2(unit, true);
		case 5: return 2 * getSum2(unit, highIsGood) + 1 * getSum2(unit, true) + (unit == 1);
		case 6: return 2 * getSum2(unit, highIsGood) + 2 * getSum2(unit, true) + (unit == 1);
		case 7: return 2 * getSum2(unit, highIsGood) + 3 * getSum2(unit, true);
		case 8: return 2 * getSum2(unit, highIsGood) + 3 * getSum2(unit, true) + highIsGood*(unit == 1);
		case 9: return 3 * getSum2(unit, highIsGood) + 3 * getSum2(unit, true) + (unit == 1);
		}
	}

	int getSum2(int unit, bool highIsGood){
		if (1 == unit) 
			return highIsGood ? 1 : 0;

		if (highIsGood) return 7 * getSum2(unit / 10, true);
			return 4 * getSum2(unit / 10, true) + 3 * getSum2(unit / 10, false);
	}
};