a***e 发帖数: 413 | 1 这些题看着容易,老是写不对或者代码很冗长,有大牛们指点一下如何提高么?
呜呜,不知道哪年哪月才能刷完一遍,有同学互勉一下么?
比如
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.
我的
Runtime Error
Last executed input:
{1}, 0
ListNode *partition(ListNode *head, int x) {
if(head==NULL)
return head;
ListNode *dummy = new ListNode(0);
dummy->next = head;
ListNode *cur = head->next;
ListNode *curHead = head;
ListNode *prev = head;
bool flag = false;
while(curHead!=NULL||cur!=NULL||prev!=NULL){
if ((cur->valval>x)){
cur = cur?cur->next:NULL;
curHead = curHead->next;
prev = prev->next;
}
else if (flag==false&&cur->val>=x){
flag=true;
prev = prev->next;
cur = cur?cur->next:NULL;
}
else if (flag==true&&cur->val
prev->next = cur->next;
cur->next = curHead->next;
curHead->next = cur;
curHead = cur;
}
else{
cur = cur?cur->next:NULL;
curHead = curHead->next;
prev = prev->next;
}
}
return dummy->next;
}
看到别的答案,
ListNode* partition(ListNode* head, int x) {
ListNode left_dummy(-1); //
ListNode right_dummy(-1); //
auto left_cur = &left_dummy;
auto right_cur = &right_dummy;
for (ListNode *cur = head; cur; cur = cur->next) {
if (cur->val < x) {
left_cur->next = cur;
left_cur = cur;
} else {
right_cur->next = cur;
right_cur = cur;
}
}
left_cur->next = right_dummy.next;
right_cur->next = nullptr;
return left_dummy.next; |
r*******k 发帖数: 1423 | 2 扫描一遍链表
整两个新链表
每个点跟x比,比x大放一个,小于放另一个
最后把这两个链表接起来就行了
记得要把拿出来的node的next置成null
否则容易有环
the
【在 a***e 的大作中提到】 : 这些题看着容易,老是写不对或者代码很冗长,有大牛们指点一下如何提高么? : 呜呜,不知道哪年哪月才能刷完一遍,有同学互勉一下么? : 比如 : 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.
|
M**a 发帖数: 848 | |
r*******k 发帖数: 1423 | 4 同感
【在 M**a 的大作中提到】 : 已经刷完了。可是很多又忘了。
|
h*******e 发帖数: 1377 | 5 这就达到张三丰教张无忌耍剑的境界了。
【在 M**a 的大作中提到】 : 已经刷完了。可是很多又忘了。
|
g*********5 发帖数: 1145 | 6 非常艰难的刷了20几道题,早忘了,更别提后面的没做过的了 |
a***n 发帖数: 623 | 7 链表题目就是考细心、考corner case、考指针来回穿插衔接的。静下心做就好。 |
y***n 发帖数: 1594 | 8 数据结构都还好,最搞的就是Array的题,变化太多,有时候很难想到好的答案。。 |
a***e 发帖数: 413 | 9 多谢各位,但现在试着在纸上跑程序,但有些就是搞不清哪里错了,比如这个
Copy List with Random Pointer
A linked list is given such that each node contains an additional random
pointer which could point to
any node in the list or null.
Return a deep copy of the list.
我看了idea,自己写了如下程序,但就是
Submission Result: Runtime Error
Last executed input:
{-1,#}
看了半天,试着一步一步走进去,还是不知道哪里错了,打算在VS里面debug一下看。
觉得和能通过的答案没有多少区别啊。。。。。。。。
My wrong answer.
/**
* Definition for singly-linked list with a random pointer.
* struct RandomListNode {
* int label;
* RandomListNode *next, *random;
* RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
* };
*/
class Solution {
public:
RandomListNode *copyRandomList(RandomListNode *head) {
if (head==NULL) return NULL;
RandomListNode *cur = head;
while(cur!=NULL){
RandomListNode *copied = new RandomListNode(cur->label);
copied->next = cur->next;
cur->next = copied;
cur = copied->next;
}
cur=head;
while(cur!=NULL){
if(cur->random!=NULL){
cur->next->random = cur->random->next;
}
cur = cur->next->next;
}
RandomListNode *newHead = head->next;
cur = newHead;
RandomListNode *oc = head;
while(cur){
oc->next = oc->next->next;
oc = oc->next;
if (cur->next){
cur->next = cur->next->next;
cur = cur->next;
}
}
return newHead;
}
};
Others' correct answer
for (RandomListNode* cur = head; cur != nullptr; ) {
RandomListNode* node = new RandomListNode(cur->label);
node->next = cur->next;
cur->next = node;
cur = node->next;
}
for (RandomListNode* cur = head; cur != nullptr; ) {
if (cur->random != NULL)
cur->next->random = cur->random->next;
cur = cur->next->next;
}
//
RandomListNode dummy(-1);
for (RandomListNode* cur = head, *new_cur = &dummy;
cur != nullptr; ) {
new_cur->next = cur->next;
new_cur = new_cur->next;
cur->next = cur->next->next;
cur = cur->next;
}
return dummy.next; |
a***e 发帖数: 413 | 10 多谢各位,但现在试着在纸上跑程序,但有些就是搞不清哪里错了,比如这个
Copy List with Random Pointer
A linked list is given such that each node contains an additional random
pointer which could point to
any node in the list or null.
Return a deep copy of the list.
我看了idea,自己写了如下程序,但就是
Submission Result: Runtime Error
Last executed input:
{-1,#}
看了半天,试着一步一步走进去,还是不知道哪里错了,打算在VS里面debug一下看。
觉得和能通过的答案没有多少区别啊。。。。。。。。
My wrong answer.
/**
* Definition for singly-linked list with a random pointer.
* struct RandomListNode {
* int label;
* RandomListNode *next, *random;
* RandomListNode(int x) : label(x), next(NULL), random(NULL) {}
* };
*/
class Solution {
public:
RandomListNode *copyRandomList(RandomListNode *head) {
if (head==NULL) return NULL;
RandomListNode *cur = head;
while(cur!=NULL){
RandomListNode *copied = new RandomListNode(cur->label);
copied->next = cur->next;
cur->next = copied;
cur = copied->next;
}
cur=head;
while(cur!=NULL){
if(cur->random!=NULL){
cur->next->random = cur->random->next;
}
cur = cur->next->next;
}
RandomListNode *newHead = head->next;
cur = newHead;
RandomListNode *oc = head;
while(cur){
oc->next = oc->next->next;
oc = oc->next;
if (cur->next){
cur->next = cur->next->next;
cur = cur->next;
}
}
return newHead;
}
};
Others' correct answer
for (RandomListNode* cur = head; cur != nullptr; ) {
RandomListNode* node = new RandomListNode(cur->label);
node->next = cur->next;
cur->next = node;
cur = node->next;
}
for (RandomListNode* cur = head; cur != nullptr; ) {
if (cur->random != NULL)
cur->next->random = cur->random->next;
cur = cur->next->next;
}
//
RandomListNode dummy(-1);
for (RandomListNode* cur = head, *new_cur = &dummy;
cur != nullptr; ) {
new_cur->next = cur->next;
new_cur = new_cur->next;
cur->next = cur->next->next;
cur = cur->next;
}
return dummy.next; |
|
|
a***e 发帖数: 413 | 11 Ft, 用VS不到半分钟就发现问题。。。。。。。。。。。。
// while(cur){
while(cur&&oc){ //加这个check就行了
oc->next = oc->next->next;
oc = oc->next;
if (cur->next){
cur->next = cur->next->next;
cur = cur->next;
}
}
但这段程序就不用那么多check
//
RandomListNode dummy(-1);
for (RandomListNode* cur = head, *new_cur = &dummy;
cur != nullptr; ) {
new_cur->next = cur->next;
new_cur = new_cur->next;
cur->next = cur->next->next;
cur = cur->next;
} |
i*****t 发帖数: 68 | 12 画图,dummy头指针很handy,大多数用recursion和双指针 |
p*******u 发帖数: 949 | |
m******3 发帖数: 346 | 14 runtime error一般都是访问了不该访问的memory,比如数组越界,指针访问错误, 你
自己把程序拿出来在自己机器上跑跑看
【在 a***e 的大作中提到】 : 多谢各位,但现在试着在纸上跑程序,但有些就是搞不清哪里错了,比如这个 : Copy List with Random Pointer : : A linked list is given such that each node contains an additional random : pointer which could point to : any node in the list or null. : Return a deep copy of the list. : : 我看了idea,自己写了如下程序,但就是 : Submission Result: Runtime Error
|
l***i 发帖数: 1309 | 15 推荐Sedgewick的algorithm in C,看相应章节,理解他的写法就好了。 |