优先队列

以后同步到csdn中

以下写在https://blog.csdn.net/ycllycll/article/details/102238286中

leetcode322  用BFS 动态规划两种解法写在 279 下面（他们两思想一样）

127. Word Ladder（也是转化为无权图的最短路径-bfs）

Given two words (beginWord and endWord), and a dictionary's word list, find the length of shortest transformation sequence from beginWord to endWord, such that:

1. Only one letter can be changed at a time.
2. Each transformed word must exist in the word list. Note that beginWord is nota transformed word.

Input:
beginWord = "hit",
endWord = "cog",
wordList = ["hot","dot","dog","lot","log","cog"]

Output: 5

Explanation: As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length 5.

代码如下：

public class Solution {
    public int ladderLength(String beginWord, String endWord, List<String> wordList) {
        Set<String> dict = new HashSet<>(wordList);
        Set<String> vis = new HashSet<>();
        Queue<String> q = new LinkedList<>();
        q.offer(beginWord);
        for (int len = 1; !q.isEmpty(); len++) {
            for (int i = q.size(); i > 0; i--) {
                String w = q.poll();
                if (w.equals(endWord)) return len;

                for (int j = 0; j < w.length(); j++) {
                    char[] ch = w.toCharArray();
                    for (char c = 'a'; c <= 'z'; c++) {
                        if (c == w.charAt(j)) continue;
                        ch[j] = c;
                        String nb = String.valueOf(ch);
                        if (dict.contains(nb) && vis.add(nb)) q.offer(nb);
                    }
                }
            }
        }
        return 0;
    }
}

126. Word Ladder II  比较难

  
一、优先队列
 java的PriorityQueue默认是一个最小堆。

PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>(); //小顶堆，默认容量为11
PriorityQueue<Integer> maxHeap = new PriorityQueue<Integer>(11,new Comparator<Integer>(){ //大顶堆，容量11
    @Override
    public int compare(Integer i1,Integer i2){
        return i2-i1;
    }
});

347. Top K Frequent Elements

Given a non-empty array of integers, return the k most frequent elements.

Example 1:

Input: nums = [1,1,1,2,2,3], k = 2
Output: [1,2]

Example 2:

Input: nums = [1], k = 1
Output: [1]

//方法一 use maxHeap. Put entry into maxHeap so we can always poll a number with largest frequency
public class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        for(int n: nums){
            map.put(n, map.getOrDefault(n,0)+1);
        }
           
        PriorityQueue<Map.Entry<Integer, Integer>> maxHeap = 
                         new PriorityQueue<>((a,b)->(b.getValue()-a.getValue()));//安对象的value来建立大顶堆
        for(Map.Entry<Integer,Integer> entry: map.entrySet()){
            maxHeap.add(entry);
        }
        
        List<Integer> res = new ArrayList<>();
        while(res.size()<k){
            Map.Entry<Integer, Integer> entry = maxHeap.poll();
            res.add(entry.getKey());
        }
        return res;
    }
}





//方法二 桶排序
public List<Integer> topKFrequent(int[] nums, int k) {

    List<Integer>[] bucket = new List[nums.length + 1];
    Map<Integer, Integer> frequencyMap = new HashMap<Integer, Integer>();

    for (int n : nums) {
        frequencyMap.put(n, frequencyMap.getOrDefault(n, 0) + 1);
    }

    for (int key : frequencyMap.keySet()) {
        int frequency = frequencyMap.get(key);
        if (bucket[frequency] == null) {
            bucket[frequency] = new ArrayList<>();
        }
        bucket[frequency].add(key);
    }

    List<Integer> res = new ArrayList<>();

    for (int pos = bucket.length - 1; pos >= 0 && res.size() < k; pos--) {
        if (bucket[pos] != null) {
            res.addAll(bucket[pos]);
        }
    }
    return res;
}

23. Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Example:

Input:
[
  1->4->5,
  1->3->4,
  2->6
]
Output: 1->1->2->3->4->4->5->6

代码如下：

//使用优先队列
public class Solution {
    public ListNode mergeKLists(List<ListNode> lists) {
        if (lists==null||lists.size()==0) return null;
        
        PriorityQueue<ListNode> queue= new PriorityQueue<ListNode>(lists.size(),new Comparator<ListNode>(){
            @Override
            public int compare(ListNode o1,ListNode o2){
                if (o1.val<o2.val)
                    return -1;
                else if (o1.val==o2.val)
                    return 0;
                else 
                    return 1;
            }
        });
        
        ListNode dummy = new ListNode(0);
        ListNode tail=dummy;
        
        for (ListNode node:lists)
            if (node!=null)
                queue.add(node);
            
        while (!queue.isEmpty()){
            tail.next=queue.poll();
            tail=tail.next;
            
            if (tail.next!=null)
                queue.add(tail.next);
        }
        return dummy.next;
    }
}

//也可以使用归并排序