简单题
21. 合并两个有序链表（无题解）
23. 合并K个升序链表（无题解）
148. 排序链表（无题解）
234. 回文链表（无题解）
剑指 Offer 24. 反转链表（无题解）
剑指 Offer 18. 删除链表的节点（无题解）
876. 链表的中间结点（无题解）
中等题
难题
题解
struct TreeNode* hash[100010];int hashsize = 0;
void visit(struct TreeNode * root){
if(root){
if(root->left) visit(root->left);
hash[hashsize++] = root;
if(root->right) visit(root->right);
}
}
struct TreeNode* convertBiNode(struct TreeNode* root){
hashsize = 0;
visit(root);
for(int i = 0;i < hashsize - 1;i++){
hash[i]->left =0;
hash[i]->right = hash[i+1];
}
if(!hashsize) return NULL;
hash[hashsize - 1]->left = 0;
hash[hashsize - 1]->right = 0;
return hash[0];
}
struct ListNode* removeDuplicateNodes(struct ListNode* head){
int hash[20001];
memset(hash,0,sizeof(hash));
struct ListNode anshead;
anshead.next = head;head = &anshead;//头节点
while(head->next){
if(!hash[head->next->val]){
hash[head->next->val] ++;
head = head->next;
}
else{
head ->next= head->next->next;
}
}
return anshead.next;
}
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
int jinwei = 0;
struct ListNode *head = malloc(sizeof(struct ListNode));
struct ListNode *ans = head;//很多地方用头节点很省事 就 很省事？
head->next = 0; //创建头节点
while(l1&&l2){
ans->next = malloc(sizeof(struct ListNode)); //创建节点
ans = ans->next;
ans->val = (l1->val + l2->val + jinwei) % 10;//创建信息
jinwei = (l1->val + l2->val + jinwei) / 10;
l1=l1->next;
l2=l2->next;
}
while(l1){ //处理l1剩余
ans->next = malloc(sizeof(struct ListNode)); //创建节点
ans = ans->next;
ans->val = (l1->val + jinwei) % 10;//创建信息
jinwei = (l1->val + jinwei) / 10;
l1=l1->next;
}
while(l2){ //处理l2剩余
ans->next = malloc(sizeof(struct ListNode)); //创建节点
ans = ans->next;
ans->val = (l2->val + jinwei) % 10;//创建信息
jinwei = (l2->val + jinwei) / 10;
l2=l2->next;
}
while(jinwei){ //处理进位剩余
ans->next = malloc(sizeof(struct ListNode)); //创建节点
ans = ans->next;
ans->val = jinwei % 10;
jinwei/= 10;
}
ans->next = 0;
return head->next;
}
struct ListNode* partition(struct ListNode* head, int x){
struct ListNode *small = malloc(sizeof(struct ListNode)),*p = small;//两个链表分别执行尾插
struct ListNode *big = malloc(sizeof(struct ListNode)),*q = big;
big->next = 0;//为了处理空串
//根据元素进行插入
while(head){
if(head->val < x){
p->next = head;
p = p->next;
}
else{
q->next = head;
q = q->next;
}
head = head->next;
}
//链接链表
p->next = big->next;
q->next = 0;
return small->next;
}
struct TreeNode *zhan[4090];//保存先序遍历的顺序
int zhansize = 0;
void first_root(struct TreeNode * root){
if(root){
zhan[zhansize++] = root;
if(root->left) first_root(root->left);
if(root->right) first_root(root->right);
}
}
void flatten(struct TreeNode* root){
first_root(root);
int i;
for(i = 0;i <zhansize - 1;i++){
zhan[i]->left = 0;
zhan[i]->right = zhan[i + 1];
}
if(!zhansize) return;
zhan[i]->left = 0;
zhan[i]->right = 0;
return;
}
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2){
struct ListNode head,*p=&head;
int jinwei = 0;
while(l1&&l2){
struct ListNode *q = malloc(sizeof(struct ListNode));
q->val = (l1->val + l2->val +jinwei) % 10;
jinwei = (l1->val + l2->val +jinwei) / 10;
p->next = q;
p = q;
l1 = l1->next;
l2 = l2->next;
}
while(l1){
struct ListNode *q = malloc(sizeof(struct ListNode));
q->val = (l1->val +jinwei) % 10;
jinwei = (l1->val +jinwei) / 10;
p->next = q;
p = q;
l1 = l1->next;
}
while(l2){
struct ListNode *q = malloc(sizeof(struct ListNode));
q->val = (l2->val +jinwei) % 10;
jinwei = (l2->val +jinwei) / 10;
p->next = q;
p = q;
l2 = l2->next;
}
while(jinwei){
struct ListNode *q = malloc(sizeof(struct ListNode));
q->val = (jinwei) % 10;
jinwei /= 10;
p->next = q;
p = q;
}
p->next = 0;
return head.next;
}
struct Node* copyRandomList(struct Node* head) {
if(!head) return head;
struct Node * q = head;
while(q){ //在每个head节点后面复制一个相同的链表
struct Node *p = malloc(sizeof(struct Node));
p->val = q->val;
p->next = q->next;
q->next = p;
q = p->next;
}
q = head;
while(q){ //复制random的指向 因为每个对应节点都在原节点后面
if(q->random) q->next->random = q->random->next;
else q->next->random = 0;
q = q->next->next;
}
q = head->next; //拆分出结果
while(q->next){
q ->next = q->next->next;
q = q->next;
}
return head->next;
}
struct ListNode* insertionSortList(struct ListNode* head){
if(!head->next||!head) return head;
struct ListNode *p = head->next;
struct ListNode anshead;//头指针 反正不返回 就在栈区把~~
anshead.next = head; //头指针便于插入
head = &anshead;
head->next->next = NULL; //分表
while(p){
struct ListNode *q = head,*temp = p->next;
while(q->next && q->next->val < p->val) q = q->next;
p->next = q->next;
q->next = p;
p = temp;
}
return head->next;
}
struct ListNode* oddEvenList(struct ListNode* head){
struct ListNode *p=0,*q=0,*ji=0,*ou=0;
int flag = 0,count = 0;
while(head){ //遍历链表
if(count&1){ //奇数
if(!flag) flag = 1;
if(!ji)
ji = head,p = ji;
else
p->next = head,p = head;
}
else {
if(!flag) flag = 2;
if(!ou)
ou = head,q = ou;
else
q->next = head,q = head;
}
count ++;
head = head->next;
}
if(flag == 1) { //链接
p->next = ou;
if(ou) q->next = 0;
return ji;
}
else if(flag == 2){ //连接
q->next = ji;
if(ji) p->next = 0;
return ou;
}
else return 0; //空串
}