[Swift]LeetCode991. 壞了的計算器 | Broken Calculator

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➤微信公衆號：山青詠芝（shanqingyongzhi）
➤博客園地址：山青詠芝（https://www.cnblogs.com/strengthen/）
➤GitHub地址：https://github.com/strengthen/LeetCode
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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

---

在顯示着數字的壞計算器上，咱們能夠執行如下兩種操做：

• 雙倍（Double）：將顯示屏上的數字乘 2；
• 遞減（Decrement）：將顯示屏上的數字減 1 。

最初，計算器顯示數字 X。

返回顯示數字 Y 所需的最小操做數。

示例 1：

輸入：X = 2, Y = 3
輸出：2
解釋：先進行雙倍運算，而後再進行遞減運算 {2 -> 4 -> 3}.

示例 2：

輸入：X = 5, Y = 8
輸出：2
解釋：先遞減，再雙倍 {5 -> 4 -> 8}.

示例 3：

輸入：X = 3, Y = 10
輸出：3
解釋：先雙倍，而後遞減，再雙倍 {3 -> 6 -> 5 -> 10}.

示例 4：

輸入：X = 1024, Y = 1
輸出：1023
解釋：執行遞減運算 1023 次

提示：

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

---

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 }

---

 8ms

 1 class Solution {
 2     func brokenCalc(_ X: Int, _ Y: Int) -> Int {
 3         if X >= Y {
 4             return X - Y
 5         }
 6         if Y % 2 == 0 {
 7             return 1 + brokenCalc(X, Y / 2)
 8         } else {
 9             return 1 + brokenCalc(X, Y + 1)
10         }
11     }
12 }