# 991. Broken Calculator 坏了的计算器

@TOC

## # 题目描述

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`

1. 翻倍
2. 减1

## # 解题方法

1. 首先我们发现x要么乘2要么减1，如果x初始就比y大，那么只能一直做减法！

2. 在x小于y的情况下：

• 如果y是奇数，那么最后一个操作一定是减1（显然）
• 如果y是偶数呢？最后操作一定是乘2吗？答案是yes!

python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 X < Y:
if Y % 2 == 1:
Y += 1
res += 1
Y //= 2
res += 1
return res + X - Y
``````

## # 日期

2019 年 2 月 21 日 —— 一放假就再难抓紧了