LeetCode 818. Race Car - 码农教程

原题链接在这里：https://leetcode.com/problems/race-car/

题目：

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 length of 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.

Note:

1 <= target <= 10000.

题解：

Use BFS to find out shortest sequence.

Each step, there are 2 possibilities, either A or R. Maintain a set to store visited state.

If current position is alrady beyond 2 * target, it can't make shortest path.

Time Complexity: O(nlogn). n = target. totoally there are 2*n positions, each position could be visited 2*logn times at most. Since the speed could not be beyond logn.

Space: O(n).

AC Java:

 1 class Solution {
 2     public int racecar(int target) {
 3         if(target == 0){
 4             return 0;
 5         }
 6         
 7         int step = 0;
 8         LinkedList<int []> que = new LinkedList<>();
 9         que.add(new int[]{0, 1});
10         int curCount = 1;
11         int nextCount = 0;
12         
13         HashSet<String> visited = new HashSet<>();
14         visited.add(0 + "," + 1);
15         
16         while(!que.isEmpty()){
17             int [] cur = que.poll();
18             curCount--;
19             
20             if(cur[0] == target){
21                 return step;
22             }
23             
24             int [] next1 = new int[]{cur[0]+cur[1], 2*cur[1]};
25             int [] next2 = new int[]{cur[0], cur[1] > 0 ? -1 : 1};
26             if(0<next1[0] && next1[0]<=target*2 && !visited.contains(next1[0] + "," + next1[1])){
27                 que.add(next1);
28                 visited.add(next1[0] + "," + next1[1]);
29                 nextCount++;
30             }
31             
32             if(!visited.contains(next2[0] + "," + next2[1])){
33                 que.add(next2);
34                 visited.add(next2[0] + "," + next2[1]);
35                 nextCount++;
36             }
37             
38             if(curCount == 0){
39                 curCount = nextCount;
40                 nextCount = 0;
41                 step++;
42             }
43         }
44         
45         return -1;
46     }
47 }

原文地址：https://www.cnblogs.com/Dylan-Java-NYC/p/11531879.html