[Leetcode Learning-c++&java] Broken Calculator (limited operation)

problem:

Difficulty: medium

Description:

Give the number X, then change X to Y, where X operates:

X can be multiplied by 2

X can be subtracted by 1.

Title link: https://leetcode.com/problems/broken-calculator/

Input range:

1. 1 <= X <= 10^9
2. 1 <= Y <= 10^9

 

Enter the case:

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.

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My code:

In fact, it is not difficult, but you need to think backwards, otherwise it will be overtime if you use dfs. If you change X to Y in the forward direction, you will be faced with the choice of * 2 or -1, and dfs is not suitable.

So the reverse thinking is that Y becomes X, and then the rules are reversed

Y can be divided by 2, only when Y is even and Y> X

Y can be + 1; this way it is easy to solve.

If X becomes Y, it’s really not clear* 2 is suitable or not, but on the other hand, Y / 2 is very clear

Java：

class Solution {
    public int brokenCalc(int X, int Y) {
        int opt = 0;
        for(; X != Y; opt ++) {
            if(Y > X && (Y & 1) == 0) Y >>= 1; // 如果是偶数先除以2为快
            else if(Y > X) Y ++;  // 如果奇数且 > X，加一
            else return opt + X - Y; // 如果已经小于 X ，剩下也只有加一操作
        }
        return opt;
    }
}

C++ (actually the same):

class Solution {
public:
    int brokenCalc(int X, int Y) {
        int opt = 0;
        for(; X != Y; opt ++) {
            if(Y > X && (Y & 1) == 0) Y >>= 1; // 如果是偶数先除以2为快
            else if(Y > X) Y ++;  // 如果奇数且 > X，加一
            else return opt + X - Y; // 如果已经小于 X ，剩下也只有加一操作
        }
        return opt;
    }
};