L2-032 彩虹瓶 java
时间: 2024-12-31 20:25:57 浏览: 10
### L2-032 彩虹瓶 Java 实现
为了处理彩虹瓶问题,在Java中可以采用数组来模拟瓶子中的小球分布情况。通过读取输入数据并将其存储在一个字符列表或字符串中,进而利用排序算法对这些字符进行重新排列。
```java
import java.util.*;
public class RainbowBottle {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
String input = scanner.nextLine();
char[] balls = input.toCharArray();
int n = Integer.parseInt(scanner.nextLine());
List<Integer> operations = new ArrayList<>();
for (int i = 0; i < n; ++i) {
operations.add(scanner.nextInt() - 1); // 调整为基于零索引
}
Map<Character, Queue<Integer>> ballPositions = new HashMap<>();
for (char c : "RBVG".toCharArray()) { // 初始化队列用于记录各色球的位置
ballPositions.put(c, new LinkedList<>());
}
for (int i = 0; i < balls.length; ++i) {
ballPositions.get(balls[i]).add(i);
}
StringBuilder result = new StringBuilder(input);
for (Integer op : operations) {
Character currentBallColor = balls[op];
Queue<Integer> positions = ballPositions.get(currentBallColor);
while (!positions.isEmpty()) {
int pos = positions.poll();
if (pos != op && !ballPositions.get('R').isEmpty()) {
swap(result, pos, ballPositions.get('R').poll());
} else if (pos != op && !ballPositions.get('B').isEmpty()){
swap(result, pos, ballPositions.get('B').poll());
} else if (pos != op && !ballPositions.get('V').isEmpty()){
swap(result, pos, ballPositions.get('V').poll());
} else if (pos != op && !ballPositions.get('G').isEmpty()){
swap(result, pos, ballPositions.get('G').poll());
}
}
}
System.out.println(result.toString());
scanner.close();
}
private static void swap(StringBuilder sb, int indexA, int indexB){
char temp = sb.charAt(indexA);
sb.setCharAt(indexA, sb.charAt(indexB));
sb.setCharAt(indexB, temp);
}
}
```
此程序首先接收一组表示初始状态的字符串以及一系列操作指令。对于每一个指定的操作位置,找到对应颜色的所有实例,并尝试与其他未被移动过的不同颜色交换位置直到完成所有给定的操作[^1]。
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