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
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; } };
原文地址:https://www.cnblogs.com/yinghualuowu/p/10360603.html
时间: 2024-10-21 09:36:08