[Swift Weekly Contest 123]LeetCode991. 坏了的计算器 | Broken Calculator

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

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在显示着数字的坏计算器上，我们可以执行以下两种操作：

• 双倍（Double）：将显示屏上的数字乘 2；
• 递减（Decrement）：将显示屏上的数字减 1 。

最初，计算器显示数字 X。

返回显示数字 Y 所需的最小操作数。

示例 1：

输入：X = 2, Y = 3
输出：2
解释：先进行双倍运算，然后再进行递减运算 {2 -> 4 -> 3}.

示例 2：

输入：X = 5, Y = 8
输出：2
解释：先递减，再双倍 {5 -> 4 -> 8}.

示例 3：

输入：X = 3, Y = 10
输出：3
解释：先双倍，然后递减，再双倍 {3 -> 6 -> 5 -> 10}.

示例 4：

输入：X = 1024, Y = 1
输出：1023
解释：执行递减运算 1023 次

提示：

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

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Runtime: 8 ms

Memory Usage: 3.9 MB

 1 class Solution {
 2     func brokenCalc(_ X: Int, _ Y: Int) -> Int {
 3         var ret:Int = Int.max
 4         for d in 0...30
 5         {
 6             var t = (X<<d) - Y
 7             if t < 0 {continue}
 8             var num:Int = d
 9             for e in 0..<d
10             {
11                 num += t&1
12                 t >>= 1;
13             }
14             num += t
15             ret = min(ret, num)
16         }
17         return ret
18     }
19 }