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  • 【leetcode】991. Broken Calculator

    题目如下:

    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

    解题思路:要增加X只能做乘法操作,要减小X只能做减法。如果X>Y的话,那么只需要一直对X做减法操作直到X=Y为止,operation的次数是X-Y;对于X<Y的情况,我们可以由Y的值反推X,即如果Y是偶数,那么令Y=Y/2,如果Y是奇数,令Y=Y-1,直至Y=X为止。

    代码如下:

    class Solution(object):
        def brokenCalc(self, X, Y):
            """
            :type X: int
            :type Y: int
            :rtype: int
            """
            if X >= Y:
                return X - Y
            res = 0
            while Y != X:
                res += 1
                if Y > X and Y % 2 == 0:
                    Y = Y / 2
                else:
                    Y = Y + 1
            return res
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10365834.html
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