Day02 集合 | 30. 串联所有单词的子串、146. LRU 缓存、811. 子域名访问计数

一、30. 串联所有单词的子串

1、思路

将字符串数组存入map，key为每个元素，value为元素个数， 遍历字符串，每次截取字符串数组长度，将截取后的字符串再次分段截取并存入map1中，最后比对2个map，相同则添加索引到list中

class Solution {public List<Integer> findSubstring(String s, String[] words) {List<Integer> list = new ArrayList<>();Map<String, Integer> map = new HashMap<>();int substrLength = 0;for (String word : words) {substrLength += word.length();map.put(word, map.getOrDefault(word, 0) + 1);}for (int i = 0; i + substrLength <= s.length(); i++) {if (valid(s.substring(i, i + substrLength), words, map)) {list.add(i);}}return list;}public boolean valid(String sub, String[] words, Map<String, Integer> map) {int k = words[0].length();Map<String, Integer> map1 = new HashMap();for (int i = 0; i < sub.length(); i += k) {String key = sub.substring(i, i + k);map1.put(key, map1.getOrDefault(key, 0) + 1);}return map1.equals(map);}
}

二、146. LRU 缓存

1、思路

HashMap + 链表 ,map方便查找，链表方便头插和去尾部，以及删除

2、代码

class LRUCache {Map<Integer, Node> map = new HashMap();Node head;Node tail;int capacity;// 需要实现功能// 1、头部插入// 2、尾部删除// 3、查询数据是否存在// 4、若存在，还需要更新数据,链表需要删除，然后重新插入public LRUCache(int capacity) {this.capacity = capacity;head = new Node();tail = new Node();head.next = tail;tail.pre = head;}public int get(int key) {if (!map.containsKey(key)) {return -1;}Node node = map.get(key);remove(node);addFirst(node);return node.value;}public void put(int key, int value) {if (map.containsKey(key)) {Node node = map.get(key);// 更新node中的valuenode.value = value;remove(node);addFirst(node);// 链表更新} else {Node node = new Node();node.key = key;node.value = value;map.put(key, node);addFirst(node);if (map.size() > capacity) {map.remove(tail.pre.key);remove(tail.pre);}}}public void remove(Node node) {node.pre.next = node.next;node.next.pre = node.pre;}public void addFirst(Node node) {// head = node2// head = node =node2head.next.pre = node;node.next = head.next;head.next = node;node.pre = head;}
}class Node {Node next;Node pre;int key;int value;public Node() {}
}

三、811. 子域名访问计数

1、思路

使用hashmap存储相同数据 以及次数，使用递归拆解域名

2、代码

class Solution {public List<String> subdomainVisits(String[] cpdomains) {Map<String, Integer> map = new HashMap();List<String> list = new ArrayList();for(String cp : cpdomains){String[] split =  cp.split(" ");split(Integer.parseInt(split[0]),split[1],map);}for(String key :map.keySet()){list.add(map.get(key) +" "+ key);}return list;}public void split(int number, String domain, Map<String, Integer> map){if(map.containsKey(domain)){int value = map.get(domain);map.put(domain, number + value);}else{map.put(domain, number);}int dotIndex = domain.indexOf('.'); // 找到第一个点的位置if (dotIndex == -1) {return ; // 如果没有点，返回}split(number,domain.substring(dotIndex + 1),map); // 截取第一个点后面的内容}
}