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
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 <= X <= 10^9
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 }