题目
reverse-nodes-in-k-group
算法
* 分段反转
* 分段反转优化
* 递归
代码
* 分段反转
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if (!head || k == 1) return head;
ListNode *dummy = new ListNode(-1);
ListNode *pre = dummy, *cur = head;
dummy->next = head;
int i = 0;
while (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;
ListNode *cur = last->next;
while(cur != next) {
last->next = cur->next;
cur->next = pre->next;
pre->next = cur;
cur = last->next;
}
return last;
}
};
* 分段优化
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
ListNode *dummy = new ListNode(-1), *pre = dummy, *cur = pre;
dummy->next = head;
int num = 0;
while (cur = cur->next) ++num;
while (num >= k) {
cur = pre->next;
for (int i = 1; i < k; ++i) {
ListNode *t = cur->next;
cur->next = t->next;
t->next = pre->next;
pre->next = t;
}
pre = cur;
num -= k;
}
return dummy->next;
}
};
* 递归
class Solution {
public:
ListNode* reverseKGroup(ListNode* head, int k) {
ListNode *cur = head;
for (int i = 0; i < k; ++i) {
if (!cur) return head;
cur = cur->next;
}
ListNode *new_head = reverse(head, cur);
head->next = reverseKGroup(cur, k);
return new_head;
}
ListNode* reverse(ListNode* head, ListNode* tail) {
ListNode *pre = tail;
while (head != tail) {
ListNode *t = head->next;
head->next = pre;
pre = head;
head = t;
}
return pre;
}
};