共计 1062 个字符，预计需要花费 3 分钟才能阅读完成。
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
难度：medium
题目：在一个显示数字的坏计算器上，我们可以执行两个操作:
加倍: 将显示的数字乘以 2，或;
递减: 从显示的数字中减去 1。
最初，计算器显示的是数字 X。
返回显示数字 Y 所需的最小操作数。
思路：从Y到X，可执行的操作为自增 1 和减半。如果为奇数则自增 1, 如果为偶则减半。如果X大于Y则只能自减。
Runtime: 3 ms, faster than 100.00% of Java online submissions for Broken Calculator.Memory Usage: 36.6 MB, less than 100.00% of Java online submissions for Broken Calculator.
class Solution {
public int brokenCalc(int X, int Y) {
int step = 0;
while (X != Y) {
if (Y < X) {
step += X – Y;
break;
}
Y = Y % 2 == 1 ? Y + 1 : Y / 2;
step++;
}
return step;
}
}
