解题思路: 把1-n的值初始化大一点 保存最大最小值跟着变化最后根据下标排个序就行了
参考代码:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.StringTokenizer;
class Jgt {
public long index, key;
public Jgt(long index, long key) {
this.index = index;
this.key = key;
}
}
public class Main {
public static void main(String[] args) {
Input sc = new Input(System.in);
int n = sc.nextInt();
int m = sc.nextInt();
Jgt[] a = new Jgt[n + 1];
long Lindex = 999999999L;
long Rindex = -1;
for (int i = 1; i <= n; i++) {
a[i] = new Jgt(i, i + 10000000);
Lindex = Math.min(a[i].key, Lindex);
Rindex = Math.max(a[i].key, Rindex);
}
while (m-- > 0) {
char c = sc.next().charAt(0);
int x = sc.nextInt();
if (c == 'L') {
a[x].key = Lindex - 1;
Lindex = a[x].key;
} else {
a[x].key = Rindex + 1;
Rindex = a[x].key;
}
}
Arrays.sort(a, 1, n + 1, (o1, o2) -> (int) (o1.key - o2.key));
for (int i = 1; i <= n; i++) {
System.out.print(a[i].index + " ");
}
}
static class Input {
public BufferedReader reader;
public StringTokenizer tokenizer;
public Input(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public String nextLine() {
String str = null;
try {
str = reader.readLine();
} catch (IOException e) {
// TODO 自动生成的 catch 块
e.printStackTrace();
}
return str;
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public Double nextDouble() {
return Double.parseDouble(next());
}
public BigInteger nextBigInteger() {
return new BigInteger(next());
}
}
}0.0分
4 人评分
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