【leetcode】991. Broken Calculator

题目如下：

On a broken calculator that has a number showing on its display, we can perform two operations:

• Double: Multiply the number on the display by 2, or;
• Decrement: Subtract 1 from the number on the display.

Initially, the calculator is displaying the number X.

Return the minimum number of operations needed to display the number Y.

Example 1:

Input: X = 2, Y = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.

Example 2:

Input: X = 5, Y = 8
Output: 2
Explanation: Use decrement and then double {5 -> 4 -> 8}.

Example 3:

Input: X = 3, Y = 10
Output: 3
Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.

Example 4:

Input: X = 1024, Y = 1
Output: 1023
Explanation: Use decrement operations 1023 times.

Note:

1. 1 <= X <= 10^9
2. 1 <= Y <= 10^9

解题思路：要增加X只能做乘法操作，要减小X只能做减法。如果X>Y的话，那么只需要一直对X做减法操作直到X=Y为止，operation的次数是X-Y；对于X<Y的情况，我们可以由Y的值反推X，即如果Y是偶数，那么令Y=Y/2，如果Y是奇数，令Y=Y-1，直至Y=X为止。

代码如下：

class Solution(object):
    def brokenCalc(self, X, Y):
        """
        :type X: int
        :type Y: int
        :rtype: int
        """
        if X >= Y:
            return X - Y
        res = 0
        while Y != X:
            res += 1
            if Y > X and Y % 2 == 0:
                Y = Y / 2
            else:
                Y = Y + 1
        return res