【leetcode】链表的基本操作

题目一：

Reverse a singly linked list.
1->2->3->NULL
NULL-<1-<2-<3

解法：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* reverseList(ListNode* head) {        ListNode* pre=NULL;        ListNode* cur=head;        while(cur!=NULL)        {            ListNode* next=cur->next;            cur->next=pre;            pre=cur;            cur=next;        }        return pre;    }};

题目二：Reverse Linked List II

Reverse Linked List II

Reverse a linked list from position m to n. Do it in-place and in
one-pass.

For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note: Given m, n satisfy the following condition: 1 ≤ m ≤ n ≤ length
of list.

解法：

ListNode *reverseBetween(ListNode *head, int m, int n){    if (m == n||head==NULL)        return head;    n -= m;    //定义一个虚拟头结点，这样方便找到上一个结点    ListNode prehead(0);    prehead._next = head;    ListNode* pre = &prehead;    //找到起始结点的上一个结点    while (--m)pre = pre->_next;    ListNode *pCur = pre->_next;    while (n--)    {        ListNode *pNext= pCur->_next;        pCur->_next = pNext->_next;        pNext->_next = pre->_next;        pre->_next = pNext;    }    return prehead._next;}

题目三： Remove Duplicates from Sorted List

Remove Duplicates from Sorted List

Given a sorted linked list, delete all duplicates such that each
element appear only once.

For example, Given 1->1->2, return 1->2. Given 1->1->2->3->3, return
1->2->3.
译文：给定一个排序的链接列表，删除所有重复项，使每个元素只显示一次。

   ListNode* deleteDuplicates(ListNode* head) {    if (head == NULL||head->next==NULL)        return head;    ListNode* pre= head;    ListNode* cur = pre->next;    while (NULL!=cur)    {        //判断是否与上一个元素重复，如果重复是，进行穿针引线        if (pre->val == cur->val)        {            ListNode* del = cur;            cur= cur->next;            pre->next = cur;            delete del;            del = NULL;        }        else//继续下一次探测        {            pre = cur;            cur = cur->next;        }    }    return head;}

题目四：Partition List

Partition List
题目描述：

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x.

You should preserve the original relative order of the nodes in each of the two partitions.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

译文：给定一个链表和一个值x，对其进行分区，使得小于x的所有节点都在大于或等于x的节点之前。

您应该保留两个分区中每个分区中节点的原始相对顺序。

ListNode *partition(ListNode *head, int x){    ListNode node1(0), node2(0);//因为要分段定义了两虚拟头结点    ListNode* p1 = &node1, *p2 = &node2;    while (head)    {        if (head->val < x)        {            p1 = p1->next = head;        }        else        {            p2 = p2->next = head;        }        head = head->next;    }    p2->next=NULL;//head==NULL时，并没有把NULL赋给p2->next    p1->next = node2.next;    return node1.next;}

题目五：Add Two Numbers

dd Two Numbers
题目描述：

You are given two non-empty linked lists representing two non-negative
integers. The digits are stored in reverse order and each of their
nodes contain a single digit. Add the two numbers and return it as a
linked list.

You may assume the two numbers do not contain any leading zero, except
the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8

译文：给定两个非空的链表，表示两个非负整数。数字以相反的顺序存储，每个节点包含一个数字。求和两个数字并将其作为链表返回。

您可以假设两个数字不包含任何前导零，除了数字0本身。

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {        ListNode preHead(0), *p = &preHead;    int extra = 0;    while (l1 || l2 || extra) {        int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + extra;        extra = sum / 10;        p->next = new ListNode(sum % 10);        p = p->next;        l1 = l1 ? l1->next : l1;        l2 = l2 ? l2->next : l2;    }    return preHead.next;    }

题目六:Remove Linked List Elements

 ListNode* removeElements(ListNode* head, int val) {      ListNode** cur=&head;      while(*cur)      {          if((*cur)->val==val)          {              *cur=(*cur)->next;//修改内容          }          else          {              cur=&(*cur)->next;//修改临时变量          }      }      return head;    }

题目七：Remove Duplicates from Sorted List II

Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate
numbers, leaving only distinct numbers from the original list.

Given
1->1->1->2->3, return 2->3.

解法一：

 ListNode *deleteDuplicates(ListNode *head){    if (head == NULL || head->next == NULL)    {        return head;    }    ListNode** runner = &head;    while (*runner)    {        if ((*runner)->next && (*runner)->next->val == (*runner)->val)        {            ListNode* temp = *runner;            while (temp && (*runner)->val == temp->val)            {                temp = temp->next;            }            *runner = temp;        }        else        {            runner = &(*runner)->next;        }    }    return head;}

解法二：

ListNode* deleteDuplicates(ListNode* head) {        if (!head) return 0;        if (!head->next) return head;        int val = head->val;        ListNode* p = head->next;        if (p->val != val) {            head->next = deleteDuplicates(p);            return head;        } else {            while (p && p->val == val) p = p->next;            return deleteDuplicates(p);        }    }

题目八：Merge Two Sorted Lists

Merge Two Sorted Lists
描述：

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

解法一：

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
if (l1 == NULL)
{
return l2;
}
if (l2 == NULL)
{
return l1;
}
ListNode* ret = NULL;
if (l1->val > l2->val)
{
ret = l2;
l2 = l2->next;
}
else
{
ret = l1;
l1 = l1->next;
}
ListNode * cur = ret;
while (l1&&l2)
{
if (l1->val > l2->val)
{
cur->next = l2;
l2 = l2->next;
}
else
{
cur->next = l1;
l1 = l1->next;
}
cur = cur->next;
}
if (l1 == NULL)
{
cur->next = l2;
}
else
{
cur->next = l1;
}
return ret;
}

解法二：使用虚拟头结点，简化问题。

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2)
{
//定义一个虚拟头结点
ListNode dummy(INT_MIN);
ListNode* tail = &dummy;

while (l1&&l2){    if (l1->val < l2->val)    {        tail->next = l1;        l1 = l1->next;    }    else    {        tail->next = l2;        l2 = l2->next;    }    tail = tail->next;}tail->next = l1 ? l1 : l2;return dummy.next;

}

问题九：Swap Nodes in Pairs

Swap Nodes in Pairs
描述：

Given a linked list, swap every two adjacent nodes and return its
head.

For example, Given 1->2->3->4, you should return the list as
2->1->4->3.

Your algorithm should use only constant space. You may not modify the
values in the list, only nodes itself can be changed.

译文：

给定一个链表，交换每两个相邻的节点并返回其头。

ListNode* swapPairs(ListNode* head){    ListNode **pp = &head, *a, *b;    while ((a = *pp) && (b = a->next))    {        a->next = b->next;        b->next = a;        *pp = b;        pp = &(a->next);    }    return head;}

问题十： Reverse Nodes in k-Group

Reverse Nodes in k-Group
描述：

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the
linked list. If the number of nodes is not a multiple of k then
left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be
changed.

Only constant memory is allowed.

For example, Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5

ListNode* reverseKGroup(ListNode* head, int k){    if (head == NULL || k == 1)return head;    int num = 0;    ListNode prehead(-1);    ListNode *preheader = &prehead;    preheader->next = head;    ListNode *cur = preheader, *nex, *pre = preheader;    while (cur = cur->next)num++;    while (num >= k)    {        cur = pre->next;        nex = cur->next;        for (int i = 1; i < k; ++i)        {            cur->next = nex->next;            nex->next = pre->next;            pre->next = nex;            nex = cur->next;        }        pre = cur;        num -= k;    }    return preheader->next;}