【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 <= X <= 10^9

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

巴特西

【leetcode】991. Broken Calculator

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