leetcode 818. Race Car

Your car starts at position 0 and speed +1 on an infinite number line.  (Your car can go into negative positions.)

Your car drives automatically according to a sequence of instructions A (accelerate) and R (reverse).

When you get an instruction "A", your car does the following: position += speed, speed *= 2.

When you get an instruction "R", your car does the following: if your speed is positive then speed = -1 , otherwise speed = 1.  (Your position stays the same.)

For example, after commands "AAR", your car goes to positions 0->1->3->3, and your speed goes to 1->2->4->-1.

Now for some target position, say the lengthof the shortest sequence of instructions to get there.

Example 1:
Input: 
target = 3
Output: 2
Explanation: 
The shortest instruction sequence is "AA".
Your position goes from 0->1->3.

Example 2:
Input: 
target = 6
Output: 5
Explanation: 
The shortest instruction sequence is "AAARA".
Your position goes from 0->1->3->7->7->6.

解题思路：

这是一道很明显的BFS的题目了，但是需要进行一些剪枝，不然最后一些例子无法通过 ；

class Solution {
public:
    int racecar(int target) 
    {
        int res = 0 ;
        unordered_set<string> visited = {"0,1"} ;
        queue<pair<long , long>> q;
        q.push({0,1}) ;
        
        while(!q.empty())
        {
            for(int i = q.size() - 1 ; i >= 0 ; --i)
            {
                long pos = q.front().first , spe = q.front().second ;
                q.pop() ;
                
                if(pos == target) return res ;
                
                long new_pos = 0, new_spe = 0 ;
                string s = "" ;
                
                new_pos = pos + spe ;
                new_spe = spe * 2 ;
                s = to_string(new_pos) + "," + to_string(new_spe) ;
                
                if(new_pos >= 0 && new_pos < target * 2 && !visited.count(s))
                {
                    q.push({new_pos , new_spe}) ;
                    visited.insert(s) ;
                }
                
                new_pos = pos ;
                new_spe = spe > 0 ? -1 : 1 ;
                s = to_string(new_pos) + "," + to_string(new_spe) ;
                
                if(new_pos >= 0 && new_pos < target * 2 && !visited.count(s)) // 当new_pos >= target * 2时就不用将其放置队列中
                {
                    q.push({new_pos , new_spe}) ;
                    visited.insert(s) ;
                }
                
            }
            res++ ;
        }
        
        return -1 ;
    }
};

 

这道题还有动态规划的做法，想了半天，不会。。先mark吧 ；

来源：https://blog.csdn.net/cmy203/article/details/100079679