0991. Broken Calculator (M)

Broken Calculator (M)

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. 1 <= X <= 10^9
2. 1 <= Y <= 10^9

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题意

给定整数X和Y，可以对X进行两种操作：乘2和减1，问经过最少多少次操作能把X变为Y。

思路

正向推的话不容易，比如X=3，Y=4时，最少操作是(X-1)*2，而X=3，Y=5时，最少操作是X*2-1，每次都要考虑是先乘2还是先减1。

反向推导，Y有两种操作：除以2和加1。有以下几种情况：当Y<X时，只能加1；当Y>X时，如果Y是奇数，则只能加1，如果Y是偶数，有两种操作：除以2，或者加2后再除以2，很明显后一种操作的结果与除以2后再加1等价，但多了一步操作，所以当Y是偶数时只需要除以2。

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代码实现

Java

class Solution {
    public int brokenCalc(int X, int Y) {
        int step = 0;
        while (Y > X) {
            step++;
            if (Y % 2 == 0) {
                Y /= 2;
            } else {
                Y++;
            }
        }
        return step + X - Y;
    }
}