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  • 991. Broken Calculator

    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
    class Solution {
        public int brokenCalc(int X, int Y) {
                if(X == Y) return 0;
                if(Y < X) return X-Y;
                return 1 + (Y % 2 == 0 ? brokenCalc(X, Y/2) : brokenCalc(X, Y + 1));
            }
    }

    草了,居然可以反过来想,intuitively如何把y变成x。

    class Solution {
        public int brokenCalc(int x, int y) {
            int res = 0;
            while(y > x) {
                y = (y % 2 == 0) ? y / 2 : y + 1;
                res++;
            }
            return res + x - y;
        }
    }
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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/14426674.html
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