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 = 3Output: 2Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.

Example 2:

Input: X = 5, Y = 8Output: 2Explanation: Use decrement and then double {5 -> 4 -> 8}.

Example 3:

Input: X = 3, Y = 10Output: 3Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.

Example 4:

Input: X = 1024, Y = 1Output: 1023Explanation: Use decrement operations 1023 times.

Note:

1 <= X <= 10^91 <= 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;    }}