力扣爆刷第89天之hot100五连刷31-35

一、25. K 个一组翻转链表

题目链接：https://leetcode.cn/problems/reverse-nodes-in-k-group/description/?envType=study-plan-v2&envId=top-100-liked
思路：k个一组翻转，写好一个单独的翻转方法，然后外部控制好每次翻转的头和尾。

class Solution {public ListNode reverseKGroup(ListNode head, int k) {ListNode root = new ListNode(-1, head);ListNode p = root, fast = head;int i = 0;while(fast != null) {i++;fast = fast.next;if(i % k == 0) {p = reverse(p, fast);}}return root.next;}ListNode reverse(ListNode start, ListNode end) {ListNode p1 = start, p2 = start.next, p3 = p2;start.next = null;while(p2 != end) {ListNode t = p2.next;p2.next = p1.next;p1.next = p2;p2 = t;}p3.next = end;return p3;}}

二、138. 随机链表的复制

题目链接：https://leetcode.cn/problems/copy-list-with-random-pointer/description/?envType=study-plan-v2&envId=top-100-liked
思路：使用map做一个映射记录，key为旧节点，value为新节点，然后遍历复制新链表，random值也复制过去，然后再遍历一遍新链表，
根据key去找value，即可。

class Solution {public Node copyRandomList(Node head) {// key-旧节点 value-新节点Map<Node, Node> map = new HashMap<>();Node root = new Node(-1);Node p1 = head, p2 = root;while(p1 != null) {Node n = new Node(p1.val);n.random = p1.random;map.put(p1, n);p2.next = n;p2 = n;p1 = p1.next;}p2 = root.next;while(p2 != null) {if(p2.random != null) {p2.random = map.get(p2.random);}p2 = p2.next;}return root.next;}
}

三、148. 排序链表

题目链接：https://leetcode.cn/problems/sort-list/description/?envType=study-plan-v2&envId=top-100-liked
思路：采用归并排序，递归划分链表，直到划分到单个元素，然后再排序。

class Solution {public ListNode sortList(ListNode head) {return sortList(head, null);}public ListNode sortList(ListNode head, ListNode tail) {if (head == null) {return head;}if (head.next == tail) {head.next = null;return head;}ListNode slow = head, fast = head;while (fast != tail) {slow = slow.next;fast = fast.next;if (fast != tail) {fast = fast.next;}}ListNode mid = slow;ListNode list1 = sortList(head, mid);ListNode list2 = sortList(mid, tail);ListNode sorted = merge(list1, list2);return sorted;}public ListNode merge(ListNode head1, ListNode head2) {ListNode dummyHead = new ListNode(0);ListNode temp = dummyHead, temp1 = head1, temp2 = head2;while (temp1 != null && temp2 != null) {if (temp1.val <= temp2.val) {temp.next = temp1;temp1 = temp1.next;} else {temp.next = temp2;temp2 = temp2.next;}temp = temp.next;}if (temp1 != null) {temp.next = temp1;} else if (temp2 != null) {temp.next = temp2;}return dummyHead.next;}
}

四、23. 合并 K 个升序链表

题目链接：https://leetcode.cn/problems/merge-k-sorted-lists/description/?envType=study-plan-v2&envId=top-100-liked
思路：使用优先级队列存放链表，出队后拼接链表，然后在把下一个节点入队。

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode() {}*     ListNode(int val) { this.val = val; }*     ListNode(int val, ListNode next) { this.val = val; this.next = next; }* }*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {if(lists.length == 0) return null;PriorityQueue<ListNode> queue = new PriorityQueue<>((a, b) -> a.val - b.val);for(ListNode list : lists) {if(list != null) {queue.add(list);}}ListNode root = new ListNode();ListNode p = root;while(!queue.isEmpty()) {ListNode min = queue.poll();p.next = min;p = min;if(min.next != null) {queue.add(min.next);}}return root.next;}
}

五、146. LRU 缓存

题目链接：https://leetcode.cn/problems/lru-cache/description/?envType=study-plan-v2&envId=top-100-liked
思路：最近最少使用，get或put都会是得该元素优先级最高，容量满了就要把最久未使用的给去掉，那么可以利用LinkedHashMap的特性，底层采用哈希表和双向链表，put方法加入的元素永远在链表的最后，最先添加进来的在头结点，所以可以使用Integer first = map.keySet().iterator().next();获取头结点然后把头结点给去掉。

class LRUCache {LinkedHashMap<Integer, Integer> map;int cap = 0;public LRUCache(int capacity) {map = new LinkedHashMap<>();cap = capacity;}public int get(int key) {if (!map.containsKey(key)) {return -1;}Integer val = map.remove(key);map.put(key, val);return val;}public void put(int key, int value) {if (map.containsKey(key)) {map.remove(key);map.put(key, value);return;}if (cap <= map.size()) {Integer first = map.keySet().iterator().next();map.remove(first);}map.put(key, value);}
}