Tech in Pieces

a classic one.Two Notes:Only constant extra memory is allowed.You may not alter the values in the list’s nodes, only nodes itself may be changed.although I’ve met thing problem many times, still can’t solve it.class Solution { public ListNode reve

链表反转核心代码：while(cur.next != null){ ListNode temp = cur.next; // cur.next = temp.next; // temp.next = pre.next; // pre.next = temp;//完成上面的四步 cur每次都停留在已经被反转的部分链表的末尾}基于原理：Input: 1->2->3->4->5->NULL其实是有三个指针。

class Solution { public ListNode mergeKLists(ListNode[] lists) {//用递归去做更好 if (lists == null || lists.length == 0) return null; PriorityQueue<ListNode> queue = new PriorityQueue<>(lists.length, (a, b) -> a.val - b.val);//t

reverse nodes by K as a groupfirst, let’s try to use recursion method.the following code has two parts, reverseKGroup() and reverseLinkedList(). the first one is the recursion function, and the second one is a tool function which reverse a linkedlist by

class Solution { public ListNode sortList(ListNode head) { if(head == null || head.next == null) return head; ListNode middle = getMiddle(head); ListNode l2 = middle.next; middle.next = null;

Given a non-negative integer represented as non-empty a singly linked list of digits, plus one to the integer.Input: [1,2,3]Output: [1,2,4]idea:the one i can think of is reverse add one and reverse back.so i implemented with this:class Solution {

简单的不能再简单了 但是第一次做的时候就是死心眼 认准了只用一个指针class Solution { public ListNode removeElements(ListNode head, int val) { if (head == null) return null; ListNode dummy = new ListNode(0); dummy.next = head; ListNode pre = dum

swap in pairsclass Solution { public ListNode swapPairs(ListNode head) { if(head == null) return null; ListNode dummy = new ListNode(0); dummy.next = head; ListNode left = dummy; ListNode right = head;