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
难度:medium
题目:在一个显示数字的坏计算器上,我们可以执行两个操作:
加倍:将显示的数字乘以2,或; 递减:从显示的数字中减去1。 最初,计算器显示的是数字X。 返回显示数字Y所需的最小操作数。
思路:从Y到X,可执行的操作为自增1和减半。如果为奇数则自增1,如果为偶则减半。如果X大于Y则只能自减。
Runtime: 3 ms, faster than 100.00% of Java online submissions for Broken Calculator.
Memory Usage: 36.6 MB, less than 100.00% of Java online submissions for Broken Calculator.
class Solution { public int brokenCalc(int X, int Y) { int step = 0; while (X != Y) { if (Y < X) { step += X - Y; break; } Y = Y % 2 == 1 ? Y + 1 : Y / 2; step++; } return step; } }