思路：

简单的思维题，应该从后往前遍历操作。如果后面的对数i操作过，则前面对数i的操作都可以无视。可以通过栈这种数据结构实现后往前遍历。

AC代码：

import java.io.*;
import java.util.*;
public class Main{
    public static void main(String[] args) throws IOException {
        StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
        in.nextToken();
        int N = (int) in.nval;
        int[] order = new int[N+5];
        boolean[] visited = new boolean[N+5];
        Stack<Integer> nums = new Stack<>();
        Stack<Integer> result = new Stack<>();
        //0代表左移 1代表右移
        Stack<Integer> operate = new Stack<>();
        in.nextToken();
        int M = (int) in.nval;
        while(M--!=0){
            in.nextToken();
            if(in.sval.equals("L")){
                operate.push(0);
            }else {
                operate.push(1);
            }
            in.nextToken();
            nums.push((int) in.nval);
        }
        while(!operate.empty()){
            if(visited[nums.peek()]){
                nums.pop();
                operate.pop();
               continue;
            }
            visited[nums.peek()]=true;
            if(operate.pop()==0){
                System.out.print(nums.pop()+" ");
            }else {
                result.push(nums.pop());
            }
        }
        for (int i = 1; i <= N; i++) {
            if(!visited[i]){
                System.out.print(i+" ");
            }
        }
        while(!result.empty()){
            System.out.print(result.pop()+" ");
        }
    }
}