991.Broken-Calculator

991. Broken Calculator

题目地址

题目描述

On a broken calculator that has a number showing on its display, we can perform two operations:
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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

代码

Approach #1 Work Backwards

Time O(logY) && Space O(1)
class Solution {
public int brokenCalc(int X, int Y) {
int ans = 0;
while (Y > X) {
ans++;
if (Y % 2 == 1) {
Y++;
} else {
Y /= 2;
}
}
​
return ans + X - Y;
}
}