Reverse Nodes in k-Group
描述
Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.
If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.
You may not alter the values in the nodes, only nodes itself may be changed.
Only constant memory is allowed.
For example,
Given this linked list: 1->2->3->4->5
For k = 2, you should return: 2->1->4->3->5
For k = 3, you should return: 3->2->1->4->5
分析
无
递归版
- Java
- C++
- Python
// Reverse Nodes in k-Group
// 递归版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || head.next == null || k < 2)
return head;
ListNode next_group = head;
for (int i = 0; i < k; ++i) {
if (next_group != null)
next_group = next_group.next;
else
return head;
}
// next_group is the head of next group
// new_next_group is the new head of next group after reversion
ListNode new_next_group = reverseKGroup(next_group, k);
ListNode prev = null, cur = head;
while (cur != next_group) {
ListNode next = cur.next;
cur.next = prev != null ? prev : new_next_group;
prev = cur;
cur = next;
}
return prev; // prev will be the new head of this group
}
}
// Reverse Nodes in k-Group
// 递归版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if (head == nullptr || head->next == nullptr || k < 2)
return head;
ListNode *next_group = head;
for (int i = 0; i < k; ++i) {
if (next_group)
next_group = next_group->next;
else
return head;
}
// next_group is the head of next group
// new_next_group is the new head of next group after reversion
ListNode *new_next_group = reverseKGroup(next_group, k);
ListNode *prev = NULL, *cur = head;
while (cur != next_group) {
ListNode *next = cur->next;
cur->next = prev ? prev : new_next_group;
prev = cur;
cur = next;
}
return prev; // prev will be the new head of this group
}
};
# Reverse Nodes in k-Group
# 递归版，时间复杂度O(n)，空间复杂度O(1)
class Solution:
def reverseKGroup(self, head, k):
if head is None or head.next is None or k < 2:
return head
next_group = head
for i in range(k):
if next_group:
next_group = next_group.next
else:
return head
# next_group is the head of next group
# new_next_group is the new head of next group after reversion
new_next_group = self.reverseKGroup(next_group, k)
prev, cur = None, head
while cur != next_group:
next = cur.next
cur.next = prev if prev else new_next_group
prev = cur
cur = next
return prev # prev will be the new head of this group
迭代版
- Java
- C++
- Python
// Reverse Nodes in k-Group
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public ListNode reverseKGroup(ListNode head, int k) {
if (head == null || head.next == null || k < 2) return head;
ListNode dummy = new ListNode(-1);
dummy.next = head;
for(ListNode prev = dummy, end = head; end != null; end = prev.next) {
for (int i = 1; i < k && end != null; i++)
end = end.next;
if (end == null) break; // 不足 k 个
prev = reverse(prev, prev.next, end);
}
return dummy.next;
}
// prev 是 first 前一个元素, [begin, end] 闭区间，保证三者都不为 null
// 返回反转后的倒数第1个元素
ListNode reverse(ListNode prev, ListNode begin, ListNode end) {
ListNode end_next = end.next;
for (ListNode p = begin, cur = p.next, next = cur.next;
cur != end_next;
p = cur, cur = next, next = next != null ? next.next : null) {
cur.next = p;
}
begin.next = end_next;
prev.next = end;
return begin;
}
};
// Reverse Nodes in k-Group
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
ListNode *reverseKGroup(ListNode *head, int k) {
if (head == nullptr || head->next == nullptr || k < 2) return head;
ListNode dummy(-1);
dummy.next = head;
for(ListNode *prev = &dummy, *end = head; end; end = prev->next) {
for (int i = 1; i < k && end; i++)
end = end->next;
if (end == nullptr) break; // 不足 k 个
prev = reverse(prev, prev->next, end);
}
return dummy.next;
}
// prev 是 first 前一个元素, [begin, end] 闭区间，保证三者都不为 null
// 返回反转后的倒数第1个元素
ListNode* reverse(ListNode *prev, ListNode *begin, ListNode *end) {
ListNode *end_next = end->next;
for (ListNode *p = begin, *cur = p->next, *next = cur->next;
cur != end_next;
p = cur, cur = next, next = next ? next->next : nullptr) {
cur->next = p;
}
begin->next = end_next;
prev->next = end;
return begin;
}
};
# Reverse Nodes in k-Group
# 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution:
def reverseKGroup(self, head, k):
if head is None or head.next is None or k < 2: return head
dummy = ListNode(-1, head)
prev, end = dummy, head
while end:
i = 1
while i < k and end:
end = end.next
i += 1
if end is None:
break # 不足 k 个
prev = self.reverse(prev, prev.next, end)
end = prev.next
return dummy.next
# prev 是 first 前一个元素, [begin, end] 闭区间，保证三者都不为 null
# 返回反转后的倒数第1个元素
def reverse(self, prev, begin, end):
end_next = end.next
p = begin
cur = p.next
next = cur.next
while cur != end_next:
cur.next = p
p = cur
cur = next
next = next.next if next else None
begin.next = end_next
prev.next = end
return begin