这个博客主要是我这几天做的链表题，这两天想给自己加一个强训。这篇只有代码，没有任何讲解。下面这些代码都是在OJ上跑过的。有些代码可能比较挫。后面我看看自己能不能优化优化。由于代码没有注释，理解起来可能有些吃力，后面等我做够了题目选择典型的题和大家分享一下啊。要是对代码有疑惑的同学也可以私信我.
除了一些需要会员和题目与二叉树以及哈希相关的，下面大该是Leetcode中链表容易的部分的全部的内容了。大家可以看看
@toc
Leetcode 19. 删除链表的倒数第 N 个结点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* removeNthFromEnd(struct ListNode* head, int n)
{
if(head == NULL)
{
return NULL;
}
struct ListNode* fast = head;
struct ListNode* slow = head;
while(n--)
{
if(fast == NULL)
{
return NULL;
}
fast = fast->next;
}
struct ListNode* prev = NULL;
while(fast != NULL)
{
fast = fast->next;
prev = slow;
slow = slow->next;
}
if(prev == NULL)
{
head = head->next;
return head;
}
prev->next = slow->next;
slow->next = NULL;
free(slow);
return head;
}Leetcode 剑指 Offer 06. 从尾到头打印链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
/**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* reversePrint(struct ListNode* head, int* returnSize)
{
if(head == NULL)
{
*returnSize = 0;
return NULL;
}
int count = 0;
struct ListNode* cur = head;
while(cur != NULL)
{
cur = cur->next;
count++;
}
*returnSize = count;
int* p = (int*)malloc(sizeof(int)*count);
cur = head;
while(cur != NULL)
{
p[--count] = cur->val;
cur = cur->next;
}
return p;
}Leetcode 83. 删除排序链表中的重复元素
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* deleteDuplicates(struct ListNode* head)
{
if(head == NULL)
{
return NULL;
}
struct ListNode* prev = head;
struct ListNode* cur = head->next;
while(cur!=NULL)
{
if(cur->val != prev->val)
{
prev->next = cur;
prev = cur;
}
cur = cur->next;
}
prev->next = NULL;
return head;
}Leetcode 剑指 Offer 18. 删除链表的节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* deleteNode(struct ListNode* head, int val)
{
if(head == NULL)
{
return NULL;
}
struct ListNode* cur = head->next;
struct ListNode* prev = head;
if(head->val == val)
{
free(head);
head = cur;
return head;
}
while(cur != NULL)
{
if(cur->val == val)
{
break;
}
prev = cur;
cur = cur ->next;
}
prev->next = cur->next;
free(cur);
return head;
}Leetcode 剑指 Offer 24. 反转链表 & 剑指 Offer II 024. 反转链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* reverseList(struct ListNode* head)
{
if(head == NULL)
{
return NULL;
}
struct ListNode* newHead = NULL;
struct ListNode* cur = head;
while(cur != NULL)
{
struct ListNode* next = cur->next;
if(newHead == NULL)
{
newHead = cur;
newHead->next = NULL;
}
else
{
cur->next = newHead;
newHead = cur;
}
cur = next;
}
return newHead;
}Leetcode 剑指 Offer 25. 合并两个排序的链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2)
{
if(l1 == NULL)
{
return l2;
}
if(l2 == NULL)
{
return l1;
}
struct ListNode* newHead = NULL;
struct ListNode* tail = NULL;
while(l1 != NULL && l2 != NULL)
{
if(l1->val < l2->val)
{
if(newHead == NULL)
{
newHead = l1;
tail = newHead;
}
else
{
tail->next = l1;
tail = l1;
}
l1 = l1->next;
}
else
{
if(newHead == NULL)
{
newHead = l2;
tail = newHead;
}
else
{
tail->next = l2;
tail = l2;
}
l2 = l2->next;
}
}
if(l1 != NULL)
{
tail->next = l1;
}
if(l2 != NULL)
{
tail->next = l2;
}
return newHead;
}Leetcode 面试题 02.07. 链表相交 & 剑指 Offer 52. 两个链表的第一个公共节点 & 剑指 Offer II 023. 两个链表的第一个重合节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode *getIntersectionNode(struct ListNode *headA, struct ListNode *headB)
{
if(headA == NULL || headB == NULL)
{
return NULL;
}
int lenA = 1;
int lenB = 1;
struct ListNode * curA = headA;
struct ListNode * curB = headB;
while(curA->next != NULL)
{
lenA++;
curA = curA->next;
}
while(curB->next != NULL)
{
lenB++;
curB = curB->next;
}
struct ListNode * longNode = headA;
struct ListNode * shortNode = headB;
int len = lenA - lenB;
if(lenA < lenB)
{
longNode = headB;
shortNode = headA;
len = -len;
}
while(len--)
{
longNode = longNode->next;
}
while(longNode != NULL && shortNode != NULL)
{
if(longNode == shortNode)
{
return longNode;
}
longNode = longNode->next;
shortNode = shortNode->next;
}
return NULL;
}Leetcode 剑指 Offer II 027. 回文链表 & 234. 回文链表 & 面试题 02.06. 回文链表
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* middleNode(struct ListNode* head)
{
if (head == NULL)
{
return NULL;
}
struct ListNode* slow = head;
struct ListNode* fast = head;
while (fast != NULL && fast->next!=NULL)
{
slow = slow->next;
fast = fast->next->next;
}
return slow;
}
struct ListNode* reverseList(struct ListNode* head)
{
if(head == NULL)
{
return NULL;
}
struct ListNode* pHead = NULL;
struct ListNode* cur = head;
while(cur != NULL)
{
if(pHead == NULL)
{
pHead = cur;
cur = cur->next;
pHead->next = NULL;
continue;
}
struct ListNode* n = cur->next;
cur->next = pHead;
pHead = cur;
cur = n;
}
return pHead;
}
bool isPalindrome(struct ListNode* head)
{
struct ListNode* mid = middleNode(head);
struct ListNode* cur = reverseList(mid);
while(cur != NULL)
{
if(head->val != cur->val)
{
return false;
}
head = head->next;
cur = cur->next;
}
return true;
}Leetcode 面试题 02.02. 返回倒数第 k 个节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
int kthToLast(struct ListNode* head, int k)
{
if(head == NULL)
{
return NULL;
}
struct ListNode* fast = head;
struct ListNode* slow = head;
while(k--)
{
if(fast == NULL || k < 0)
{
return NULL;
}
fast = fast->next;
}
while(fast != NULL)
{
fast = fast->next;
slow = slow->next;
}
return slow->val;
}Leetcode 面试题 02.03. 删除中间节点 & 237. 删除链表中的节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
void deleteNode(struct ListNode* node)
{
if(node == NULL)
{
return;
}
struct ListNode* cur = node;
struct ListNode* next = cur->next;
while(next != NULL)
{
cur->val = next->val;
if(next->next == NULL)
{
cur->next = NULL;
}
cur = cur->next;
next = next->next;
}
}Leetcode 面试题 02.01. 移除重复节点
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
int isTrue(struct ListNode* head, struct ListNode* tail,int val)
{
if (head == NULL)
{
return 0;
}
struct ListNode* cur = head;
tail->next = NULL;
while (cur != NULL)
{
if (cur->val == val)
{
return 0;
}
cur = cur->next;
}
return 1;
}
struct ListNode* removeDuplicateNodes(struct ListNode* head){
if(head == NULL)
{
return NULL;
}
struct ListNode* newHead = NULL;
struct ListNode* tail = NULL;
struct ListNode* cur = head;
while(cur != NULL)
{
if(newHead == NULL)
{
newHead = cur;
tail = newHead;
}
else
{
if(isTrue(newHead, tail,cur->val) == 1)
{
tail->next = cur;
tail = cur;
}
}
cur = cur->next;
}
tail->next = NULL;
return newHead;
}
