链表Dummy Node
（一）Partition List
https://leetcode.com/problems/partition-list/description/
题目：给出一个链表和一个x值，要求返回一个顺序链表使得小于x的数在链表前面，大于等于x的数在链表后面，保证节点顺序不变。
例如：1->3->2->4->2, 3 变成：1->2->2->3->4
解答：建立两个新的左、右指针及dummy node，使用head指针遍历整个链表，遇到大于等于head的节点则放到右链表，否则放到左链表。最后将左右链表相连。
第一次犯错：忘记将又指针的尾部指向null;
代码：
class Solution {
public ListNode partition(ListNode head, int x) {
if (head == null) {
return head;
}
ListNode leftDummy = new ListNode(0);
ListNode rightDummy = new ListNode(0);
ListNode left = leftDummy;
ListNode right = rightDummy;
while (head != null) {
if (head.val >= x) {
right.next = head;
right = right.next;
} else {
left.next = head;
left = left.next;
}
head = head.next;
}
right.next = null;
left.next = rightDummy.next;
return leftDummy.next;
}
}
（二） Merge Two Sorted Lists
https://leetcode.com/problems/merge-two-sorted-lists/description/
AC！
题目：将两个顺序链表合并成一个顺序链表；
解答：依次比较两个链表里的值大小进行排列；
改进：当一个链表指针指向null，另一个链表还没时，可以直接将重新排列的链表尾指向当前指针：
if (l1 != null) {
head.next = l1;
} else {
head.next = l2;
}
代码：
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode head = dummy;
while (l2 != null && l1 != null) {
if (l1.val <= l2.val) {
head.next = l1;
l1 = l1.next;
} else {
head.next = l2;
l2 = l2.next;
}
head = head.next;
}
while (l1 != null) {
head.next = l1;
l1 = l1.next;
head = head.next;
}
while (l2 != null) {
head.next = l2;
l2 = l2.next;
head = head.next;
}
head.next = null;
return dummy.next;
}
}
（三）swap two nodes in linked list
https://leetcode.com/problems/swap-nodes-in-pairs/description/
AC！
题目：两两交换链表中节点位置。如：1->2->4->5->6 转变为: 2->1->5->4->6
解答：使用两个指针遍历链表；
代码：
class Solution {
public ListNode swapPairs(ListNode head) {
if (head == null) {
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode headNext = head.next;
ListNode headPrev = dummy;
while (head != null && headNext != null) {
headPrev.next = headNext;
head.next = headNext.next;
headNext.next = head;
headPrev = head;
head = head.next;
if (head != null) {
headNext = head.next;
}
}
return dummy.next;
}
}
（四）reorder list
https://leetcode.com/problems/reorder-list/description/
题目：给定一个单链表：L0->L1->...->Ln-1->Ln,
重新排序后为：L0->Ln->L1->Ln-1->L2->Ln-2->...
解答：
第一次犯错：将链表尾连接到头之后，忘记将尾的前一个指针指向空；
第二次犯错：忘记考虑链表头和prevTail重合的情况（偶数链表）。
代码：
class Solution {
public void reorderList(ListNode head) {
while (head != null && head.next != null) {
ListNode prevTail = head;
while (prevTail.next.next != null) {
prevTail = prevTail.next;
}
ListNode tail = prevTail.next;
if (head.next == tail) {
break;
}
tail.next = head.next;
head.next = tail;
head = head.next.next;
prevTail.next = null;
}
}
}
（四）Rotate List
https://leetcode.com/problems/rotate-list/description/
题目：将链表尾部的k个节点移到链表头部；
解答：每次将链表最后一个节点移动至链表头，移动k次；
第一次犯错：（超时）先遍历链表，得到链表长度length，循环只需执行 k%length 次；
代码：
class Solution {
public ListNode rotateRight(ListNode head, int k) {
ListNode tail = head;
int length = 0;
while (tail != null ) {
length++;
tail = tail.next;
}
if (length == 0 || length == 1) {
return head;
}
for (int i = 0; i < k % length; i++) {
ListNode preTail = head;
while (preTail != null && preTail.next != null && preTail.next.next != null) {
preTail = preTail.next;
}
tail = preTail.next;
tail.next = head;
preTail.next = null;
head = tail;
}
return head;
}
}