题目

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 <= $10^9$

解析

X变Y等价于通过加法和除法使Y变X. 贪心地执行除法操作, 能整除2就除, 不能就加一, 直到Y比X小的时候, 此时在补上 X - Y个加一操作即可.

代码

c++

class Solution {
public:
    int brokenCalc(int X, int Y) {
        int res = 0;
        while (Y > X) {
            res++;
            Y = Y & 1 ? Y + 1 : Y >> 1;
        }

        return res + X - Y;
    }
};