1.reverse-nodes-in-k-group（k组翻转链表）【hard】
给你一个链表以及一个k,将这个链表从头指针开始每k个翻转一下。链表元素个数不是k的倍数，最后剩余的不用翻转。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
/**
* @param head a ListNode
* @param k an integer
* @return a ListNode
*/
public ListNode reverseKGroup(ListNode head, int k) {
// Write your code here
if (head == null || k <= 1) {
return head;
}
ListNode dummy = new ListNode(0);
dummy.next = head;
head = dummy;
while (head.next != null) {
head = reverseNextK(head, k);
}
return dummy.next;
}
private ListNode reverseNextK(ListNode head, int k) {
ListNode node = head;
for (int i = 0; i < k; i++) {
if (node.next == null) {
return head.next;
}
node = node.next;
}
ListNode n1 = head.next;
ListNode prev = head;
ListNode curt = head.next;
for (int i = 0; i < k; i++) {
ListNode temp = curt.next;
curt.next = prev;
prev = curt;
curt = temp;
}
n1.next = curt;
head.next = prev;
return n1;
}
}
注意：n0->n1->n2->...->nk->nk+1若要翻转n1->...->nk,则n0到nk节点都会变化。 手动创建dummy node(哨兵节点)，dummy.next总是head(头结点)，最后返回dummy.next也即head节点。reverseNextK()函数分三步：1.检查是否有足够的k个节点可翻转（如果没有,返回head.next,因为此时的head节点实际已在上次操作中被翻转）2.翻转（ n1 = head.next; prev = head; curt = head.next;for (int i = 0; i < k; i++) { temp = curt.next; curt.next = prev; prev = curt; curt = temp;}）3.链接，以便继续下次翻转（n1.next = curt; head.next = prev; return n1;）
2.reverse-linked-list（翻转链表）
翻转一个链表。给出一个链表1->2->3->null，这个翻转后的链表为3->2->1->null。
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
*/
public class Solution {
/**
* @param head: The head of linked list.
* @return: The new head of reversed linked list.
*/
public ListNode reverse(ListNode head) {
// write your code here
ListNode prev = null;
ListNode curt = head;
while (curt != null) {
ListNode temp = curt.next;
curt.next = prev;
prev = curt;
curt = temp;
}
return prev;
}
}
3.partition-list（链表划分）
给定一个单链表和数值x，划分链表使得所有小于x的节点排在大于等于x的节点之前。你应该保留两部分内链表节点原有的相对顺序。
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
*/
public class Solution {
/**
* @param head: The first node of linked list.
* @param x: an integer
* @return: a ListNode
*/
public ListNode partition(ListNode head, int x) {
// write your code here
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) {
left.next = head;
left = head;
} else {
right.next = head;
right = head;
}
head = head.next;
}
left.next = rightDummy.next;
right.next = null;
return leftDummy.next;
}
}
注意：定义leftDummy和rightDummy两个哨兵节点，分别用来保存<x和≥x节点，最后进行链接即可。在left(right).next=head之后，left(right)也要=head。
4.merge-two-sorted-lists（合并两个排序链表）
将两个排序链表合并为一个新的排序链表。
/**
* Definition for ListNode.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int val) {
* this.val = val;
* this.next = null;
* }
* }
*/
public class Solution {
/**
* @param ListNode l1 is the head of the linked list
* @param ListNode l2 is the head of the linked list
* @return: ListNode head of linked list
*/
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
// write your code here
ListNode dummy = new ListNode(0);
ListNode head = dummy;
while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
head.next = l1;
l1 = l1.next;
} else {
head.next = l2;
l2 = l2.next;
}
head = head.next;
}
if (l1 != null) {
head.next = l1;
} else {
head.next = l2;
}
return dummy.next;
}
}
注意：经典合并法。当两个链表都不为空时，比较节点值的大小...当其中一个链表不为空时...