991. Broken Calculator

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/zjucor/article/details/86906387

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

思路：X到Y难，但是Y到X容易，因为每一步都是确定的选择

Y到X就是要么（1）除以2，（2）加1

如果Y<X：只能加1，如果Y>X就要分情况讨论

1. 如果Y是奇数，必然加1

2. 如果Y是偶数，必然除以2得到Y/2，（另外一种操作方式是+1,+1,/2得到Y/2-1，但是比/2,-1用的次数多，所以必然是先要/2）

class Solution(object):
    def brokenCalc(self, X, Y):
        """
        :type X: int
        :type Y: int
        :rtype: int
        """
        res=0
        while Y>X:
            if Y%2==0:
                Y=Y//2
            else:
                Y=Y+1
            res+=1
        res+=X-Y
        return res