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
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; } }