991. Broken Calculator

最新推荐文章于 2020-06-19 13:53:44 发布

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最新推荐文章于 2020-06-19 13:53:44 发布

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文章标签： java runtime

原文链接：https://segmentfault.com/a/1190000018170906

本文探讨了一个坏计算器上的数学问题，如何通过加倍或递减操作，从初始数字X达到目标数字Y，同时提供了一种高效算法实现，该算法逆向思考，从Y到X，通过自增和减半操作，寻找最小操作数。

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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;
    }
}