0991._Broken_Calculator.md

991. Broken Calculator

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/broken-calculator/

内容描述

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

解题方案

思路 1 - 时间复杂度: O(lg(Y-X))- 空间复杂度: O(1)******

soluition说的很清楚了

也可以看看[Java/C++/Python] Change Y to X in 1 Line

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

思路 2 - 时间复杂度: O(lg(Y-X))- 空间复杂度: O(1)******

递归

class Solution:
    def brokenCalc(self, X: 'int', Y: 'int') -> 'int':
        return X - Y if X >= Y else 1 + Y % 2 + self.brokenCalc(X, (Y + 1) // 2)