Rearrange String k Distance Apart

最新推荐文章于 2024-12-28 17:28:49 发布

weixin_33905756

最新推荐文章于 2024-12-28 17:28:49 发布

阅读量89

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/shuashuashua/p/5620485.html

本文介绍了一种基于优先队列的数据结构实现的字符串重排算法，该算法能够在指定间隔k的情况下重新排列字符串中的字符，确保相同的字符至少相隔k个位置。通过使用HashMap统计字符频率并借助优先队列进行高效处理，此方法适用于需要避免重复字符连续出现的场景。

 1 public class Solution {
 2     public String rearrangeString(String str, int k) {
 3         if (k <= 1) {
 4             return str;
 5         }
 6         
 7         Map<Character, Integer> letters = new HashMap<>();
 8         for (char c : str.toCharArray()) {
 9             letters.put(c, letters.getOrDefault(c, 0) + 1);
10         }
11         
12         PriorityQueue<Map.Entry<Character, Integer>> startQueue = new PriorityQueue<>(new Comparator<Map.Entry<Character, Integer>>() {
13            @Override
14            public int compare(Map.Entry<Character, Integer> m1, Map.Entry<Character, Integer> m2) {
15                return m2.getValue().compareTo(m1.getValue());
16            }
17         });
18         
19         for (Map.Entry<Character, Integer> entry: letters.entrySet()) {
20             startQueue.offer(entry);
21         }
22         
23         StringBuilder result = new StringBuilder();
24         LinkedList<Map.Entry<Character, Integer>> backupList = new LinkedList<>();
25         
26         while (!startQueue.isEmpty()) {
27             Map.Entry<Character, Integer> current = startQueue.poll();
28             result.append(current.getKey());
29             current.setValue(current.getValue() - 1);
30             backupList.offer(current);
31             
32             if (backupList.size() < k) {
33                 continue;
34             }
35             
36             current = backupList.pollFirst();
37             if (current.getValue() > 0) {
38                 startQueue.offer(current);
39             }
40         }
41         return result.length() == str.length() ? result.toString() : "";
42     }
43 }