解题思路
这是一个模拟题,需要处理以下几种情况:
-
歌曲总数 ≤ 4:不需要翻页,只移动光标
- Up键:光标在第一首时移到最后,否则向上移动
- Down键:光标在最后时移到第一首,否则向下移动
-
歌曲总数 > 4:需要处理翻页
- 特殊翻页:
- 第一页按Up:显示最后一页,光标到最后
- 最后一页按Down:显示第一页,光标到最前
- 一般翻页:
- 光标在当前页第一首按Up:向上翻页
- 光标在当前页最后一首按Down:向下翻页
- 特殊翻页:
代码
#include <iostream>
#include <vector>
using namespace std;
class MP3Player {
private:
int n; // 总歌曲数
int cursor; // 当前光标位置(1-based)
int pageStart; // 当前页面起始歌曲编号(1-based)
void handleSmallList(char cmd) {
if (cmd == 'U') {
cursor = (cursor == 1) ? n : cursor - 1;
} else {
cursor = (cursor == n) ? 1 : cursor + 1;
}
}
void handleLargeList(char cmd) {
if (cmd == 'U') {
if (cursor == 1) {
// 特殊翻页:第一首歌按Up
cursor = n;
pageStart = max(1, n - 3);
} else {
cursor--;
if (cursor < pageStart) {
pageStart = cursor;
}
}
} else {
if (cursor == n) {
// 特殊翻页:最后一首歌按Down
cursor = 1;
pageStart = 1;
} else {
cursor++;
if (cursor > pageStart + 3) {
pageStart = cursor - 3;
}
}
}
}
public:
MP3Player(int totalSongs) : n(totalSongs), cursor(1), pageStart(1) {}
void process(char cmd) {
if (n <= 4) {
handleSmallList(cmd);
} else {
handleLargeList(cmd);
}
}
void display() {
// 显示当前页面
int end = min(pageStart + 3, n);
for (int i = pageStart; i <= end; i++) {
cout << i << " ";
}
cout << endl;
// 显示当前选中歌曲
cout << cursor << endl;
}
};
int main() {
int n;
string commands;
cin >> n >> commands;
MP3Player player(n);
for (char cmd : commands) {
player.process(cmd);
}
player.display();
return 0;
}
import java.util.*;
public class Main {
static class MP3Player {
private int n; // 总歌曲数
private int cursor; // 当前光标位置(1-based)
private int pageStart; // 当前页面起始歌曲编号(1-based)
public MP3Player(int totalSongs) {
this.n = totalSongs;
this.cursor = 1;
this.pageStart = 1;
}
private void handleSmallList(char cmd) {
if (cmd == 'U') {
cursor = (cursor == 1) ? n : cursor - 1;
} else {
cursor = (cursor == n) ? 1 : cursor + 1;
}
}
private void handleLargeList(char cmd) {
if (cmd == 'U') {
if (cursor == 1) {
cursor = n;
pageStart = Math.max(1, n - 3);
} else {
cursor--;
if (cursor < pageStart) {
pageStart = cursor;
}
}
} else {
if (cursor == n) {
cursor = 1;
pageStart = 1;
} else {
cursor++;
if (cursor > pageStart + 3) {
pageStart = cursor - 3;
}
}
}
}
public void process(char cmd) {
if (n <= 4) {
handleSmallList(cmd);
} else {
handleLargeList(cmd);
}
}
public void display() {
int end = Math.min(pageStart + 3, n);
for (int i = pageStart; i <= end; i++) {
System.out.print(i + " ");
}
System.out.println();
System.out.println(cursor);
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
String commands = sc.next();
MP3Player player = new MP3Player(n);
for (char cmd : commands.toCharArray()) {
player.process(cmd);
}
player.display();
}
}
class MP3Player:
def __init__(self, total_songs):
self.n = total_songs # 总歌曲数
self.cursor = 1 # 当前光标位置(1-based)
self.page_start = 1 # 当前页面起始歌曲编号(1-based)
def handle_small_list(self, cmd):
if cmd == 'U':
self.cursor = self.n if self.cursor == 1 else self.cursor - 1
else:
self.cursor = 1 if self.cursor == self.n else self.cursor + 1
def handle_large_list(self, cmd):
if cmd == 'U':
if self.cursor == 1:
# 特殊翻页:第一首歌按Up
self.cursor = self.n
self.page_start = max(1, self.n - 3)
else:
self.cursor -= 1
if self.cursor < self.page_start:
self.page_start = self.cursor
else:
if self.cursor == self.n:
# 特殊翻页:最后一首歌按Down
self.cursor = 1
self.page_start = 1
else:
self.cursor += 1
if self.cursor > self.page_start + 3:
self.page_start = self.cursor - 3
def process(self, cmd):
if self.n <= 4:
self.handle_small_list(cmd)
else:
self.handle_large_list(cmd)
def display(self):
# 显示当前页面
end = min(self.page_start + 3, self.n)
print(' '.join(map(str, range(self.page_start, end + 1))))
# 显示当前选中歌曲
print(self.cursor)
# 处理输入
n = int(input())
commands = input().strip()
player = MP3Player(n)
for cmd in commands:
player.process(cmd)
player.display()
算法及复杂度
- 算法:模拟
- 时间复杂度:
,其中
是命令的长度
- 空间复杂度:
,只需要常数额外空间
这个解法通过模拟MP3播放器的行为,维护当前光标位置和页面起始位置,根据不同情况处理上下键的操作。代码结构清晰,分别处理了歌曲数量小于等于4和大于4的情况。