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25.cpp
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25.cpp
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__________________________________________________________________________________________________
20ms
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
ListNode *cur = head;
int i;
for (i = k; i > 0 && cur; i--) {
cur = cur->next;
}
if (i == 0) {
ListNode *pre = reverseKGroup(cur, k), *next;
cur = head;
while (k-- > 0) {
next = cur->next;
cur->next = pre;
pre = cur;
cur = next;
}
return pre;
}
return head;
}
};
__________________________________________________________________________________________________
20ms
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
if (!head || k == 1) return head;
ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = head;
dummy->next = head;
for (int i = 1; cur; ++i) {
if (i % k == 0) {
pre = reverseOneGroup(pre, cur->next);
cur = pre->next;
} else {
cur = cur->next;
}
}
return dummy->next;
}
ListNode* reverseOneGroup(ListNode* pre, ListNode* next) {
ListNode *last = pre->next, *cur = last->next;
while(cur != next) {
last->next = cur->next;
cur->next = pre->next;
pre->next = cur;
cur = last->next;
}
return last;
}
};
__________________________________________________________________________________________________
24ms
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode * reverseKGroup(ListNode* head, int k) {
ListNode before(0);
before.next = head;
stack<ListNode*> st;
ListNode* pre = &before;
while (pre && pre->next)
{
ListNode* tmp = pre;
while (st.size()<k)
{
if (pre->next) {
st.push(pre->next);
pre = pre->next;
}
else {
break;
}
}
if (st.size() == k) {
ListNode* t = tmp;
ListNode* t2 = pre->next;
while (st.size())
{
t->next = st.top();
st.pop();
t = t->next;
}
t->next = t2;
pre = t;
}
else {
break;
}
}
return before.next;
}
};
__________________________________________________________________________________________________
28ms
class Solution {
public:
/**
*pre指向可翻转的一段的前一位
*cur指向要翻转的前一个结点
*next指向要翻转的结点 头插的方式加到pre的后面
*tmp记录next的next
*之后要翻转的是tmp,所以next指向tmp
*翻转完一段后pre指向cue
*/
ListNode* reverseKGroup(ListNode* head, int k) {
if(head==NULL||head->next==NULL||k==1)
return head;
ListNode* dummy = new ListNode(-1);
dummy->next = head;
ListNode* pre = dummy;
ListNode* p = head;
int count = 0;
while(p!=NULL)
{
p = p->next;
count++;
}
while(count>=k)
{
ListNode* cur = pre->next;
ListNode* next = cur->next;
int t = k-1;
while(t--)
{
ListNode* tmp = next->next;
next->next = pre->next;
pre->next = next;
cur->next = tmp;
next = tmp;
}
pre = cur;
count-=k;
}
ListNode* res = dummy->next;
delete dummy;
return res;
}
};
__________________________________________________________________________________________________
10004 kb
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseGroup(ListNode* begin, ListNode* end){
ListNode* pre = end;
while(begin!=end){
ListNode* tmp = begin->next;
begin->next = pre;
pre = begin;
begin = tmp;
}
return pre;
}
ListNode* reverseKGroup(ListNode* head, int k) {
if (head == NULL)
return head;
int count = 0;
ListNode* begin = head;
ListNode* end = head;
while(count<k&&end!=NULL){
end = end->next;
count++;
}
if (count<k)
return head;
ListNode* newHead = reverseGroup(begin, end);
while(end!=NULL){
begin = end;
count = 0;
while(count<k&&end!=NULL){
end = end->next;
count++;
}
if (count<k)
break;
else{
head->next = reverseGroup(begin, end);
head = begin;
}
}
return newHead;
}
};
__________________________________________________________________________________________________