55 lines
1.4 KiB
C++
55 lines
1.4 KiB
C++
// Source : https://leetcode.com/problems/broken-calculator/
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// Author : Hao Chen
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// Date : 2019-05-01
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/*****************************************************************************************************
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*
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* On a broken calculator that has a number showing on its display, we can perform two operations:
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*
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* Double: Multiply the number on the display by 2, or;
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* Decrement: Subtract 1 from the number on the display.
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*
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* Initially, the calculator is displaying the number X.
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*
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* Return the minimum number of operations needed to display the number Y.
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*
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* Example 1:
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*
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* Input: X = 2, Y = 3
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* Output: 2
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* Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
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*
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* Example 2:
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*
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* Input: X = 5, Y = 8
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* Output: 2
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* Explanation: Use decrement and then double {5 -> 4 -> 8}.
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*
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* Example 3:
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*
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* Input: X = 3, Y = 10
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* Output: 3
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* Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.
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*
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* Example 4:
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*
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* Input: X = 1024, Y = 1
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* Output: 1023
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* Explanation: Use decrement operations 1023 times.
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*
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* Note:
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*
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* 1 <= X <= 10^9
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* 1 <= Y <= 10^9
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******************************************************************************************************/
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class Solution {
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public:
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int brokenCalc(int X, int Y) {
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if (X >= Y) return X-Y ;
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if ( Y%2 ==0 ) return brokenCalc(X, Y/2) + 1;
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return brokenCalc(X, Y+1) + 1;
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}
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};
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