机器人碰撞

There are n 1-indexed robots, each having a position on a line, health, and movement direction.

You are given 0-indexed integer arrays positions, healths, and a string directions (directions[i] is either 'L' for left or 'R' for right). All integers in positions are unique.

All robots start moving on the line simultaneously at the same speed in their given directions. If two robots ever share the same position while moving, they will collide.

If two robots collide, the robot with lower health is removed from the line, and the health of the other robot decreases by one. The surviving robot continues in the same direction it was going. If both robots have the same health, they are both removed from the line.

Your task is to determine the health of the robots that survive the collisions, in the same order that the robots were given, i.e. final heath of robot 1 (if survived), final health of robot 2 (if survived), and so on. If there are no survivors, return an empty array.

Return an array containing the health of the remaining robots (in the order they were given in the input), after no further collisions can occur.

Note: The positions may be unsorted.

 

Example 1:

Input: positions = [5,4,3,2,1], healths = [2,17,9,15,10], directions = "RRRRR"
Output: [2,17,9,15,10]
Explanation: No collision occurs in this example, since all robots are moving in the same direction. So, the health of the robots in order from the first robot is returned, [2, 17, 9, 15, 10].

Example 2:

Input: positions = [3,5,2,6], healths = [10,10,15,12], directions = "RLRL"
Output: [14]
Explanation: There are 2 collisions in this example. Firstly, robot 1 and robot 2 will collide, and since both have the same health, they will be removed from the line. Next, robot 3 and robot 4 will collide and since robot 4's health is smaller, it gets removed, and robot 3's health becomes 15 - 1 = 14. Only robot 3 remains, so we return [14].

Example 3:

Input: positions = [1,2,5,6], healths = [10,10,11,11], directions = "RLRL"
Output: []
Explanation: Robot 1 and robot 2 will collide and since both have the same health, they are both removed. Robot 3 and 4 will collide and since both have the same health, they are both removed. So, we return an empty array, [].

 

Constraints:

• 1 <= positions.length == healths.length == directions.length == n <= 105
• 1 <= positions[i], healths[i] <= 109
• directions[i] == 'L' or directions[i] == 'R'
• All values in positions are distinct

/*
 * 思路：
 * 1. 将机器人按位置排序。
 * 2. 使用栈来模拟机器人碰撞的过程。
 * 3. 遍历机器人，如果当前方向为'R'，将其入栈。
 * 4. 如果当前方向为'L'，则判断栈顶机器人方向是否为'R'，是则发生碰撞，根据健康度决定结果。
 * 5. 最后输出幸存机器人的健康度。
 */

import java.util.*;
import java.util.stream.*;

public class RobotSurvivorStream {
    public int[] surviveRobots(int[] positions, int[] healths, String directions) {
        int n = positions.length;
        List<Robot> robots = IntStream.range(0, n)
                .mapToObj(i -> new Robot(positions[i], healths[i], directions.charAt(i)))
                .sorted(Comparator.comparingInt(r -> r.position))
                .collect(Collectors.toList());
        Deque<Robot> stack = new ArrayDeque<>();
        robots.forEach(robot -> {
            if (robot.direction == 'R') {
                stack.push(robot);
            } else {
                while (!stack.isEmpty() && stack.peek().direction == 'R') {
                    Robot top = stack.pop();
                    if (top.health > robot.health) {
                        top.health -= 1;
                        robot = null;
                        stack.push(top);
                        break;
                    } else if (top.health < robot.health) {
                        robot.health -= 1;
                    } else {
                        robot = null;
                        break;
                    }
                }
                if (robot != null) {
                    stack.push(robot);
                }
            }
        });
        return stack.stream().sorted(Comparator.comparingInt(r -> r.position)).mapToInt(r -> r.health).toArray();
    }

    class Robot {
        int position;
        int health;
        char direction;

        Robot(int position, int health, char direction) {
            this.position = position;
            this.health = health;
            this.direction = direction;
        }
    }
}

这个算法首先通过对机器人的位置进行排序，以便按照位置顺序处理它们的移动和碰撞。使用一个栈来跟踪所有向右移动的机器人。当遇到向左移动的机器人时，检查栈顶的向右移动的机器人是否与之碰撞。如果发生碰撞，比较两个机器人的健康度，健康度较低的机器人被移除。如果健康度相同，则两者都被移除。栈用于跟踪仍然在移动的机器人，以便可以继续处理后续可能的碰撞。最后，返回所有幸存机器人的健康度。

O(n log n)

O(n)