123th LeetCode Weekly Contest 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

CF原题，我们反过来看，Y是偶数就/2，奇数就加1，到了比X小就再加

class Solution {
public:
    int brokenCalc(int X, int Y) {
        int cnt=0;
        if(X>=Y){
            return X-Y;
        }
        while(X!=Y){
            if(Y<X)Y++;
            else if(Y%2){
                Y++;
            }else{
                Y/=2;
            }
            //cout<<Y<<endl;
            cnt++;
        }
        return cnt;
    }
};

123th LeetCode Weekly Contest Broken Calculator

原文：https://www.cnblogs.com/yinghualuowu/p/10360603.html