• LeetCode 818. Race Car


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