You are given the head
of a linked list.
The nodes in the linked list are sequentially assigned to non-empty groups whose lengths form the sequence of the natural numbers (1, 2, 3, 4, ...
). The length of a group is the number of nodes assigned to it. In other words,
- The
1st
node is assigned to the first group. - The
2nd
and the3rd
nodes are assigned to the second group. - The
4th
,5th
, and6th
nodes are assigned to the third group, and so on.
Note that the length of the last group may be less than or equal to 1 + the length of the second to last group
.
Reverse the nodes in each group with an even length, and return the head
of the modified linked list.
Example 1:
Input: head = [5,2,6,3,9,1,7,3,8,4] Output: [5,6,2,3,9,1,4,8,3,7] Explanation: - The length of the first group is 1, which is odd, hence no reversal occurrs. - The length of the second group is 2, which is even, hence the nodes are reversed. - The length of the third group is 3, which is odd, hence no reversal occurrs. - The length of the last group is 4, which is even, hence the nodes are reversed.
Example 2:
Input: head = [1,1,0,6] Output: [1,0,1,6] Explanation: - The length of the first group is 1. No reversal occurrs. - The length of the second group is 2. The nodes are reversed. - The length of the last group is 1. No reversal occurrs.
Example 3:
Input: head = [2,1] Output: [2,1] Explanation: - The length of the first group is 1. No reversal occurrs. - The length of the last group is 1. No reversal occurrs.
Example 4:
Input: head = [8] Output: [8] Explanation: There is only one group whose length is 1. No reversal occurrs.
Constraints:
- The number of nodes in the list is in the range
[1, 105]
. 0 <= Node.val <= 105
Similar Questions:
// OJ: https://leetcode.com/problems/reverse-nodes-in-even-length-groups/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
int getLength(ListNode *head) {
int len = 0;
for (; head; head = head->next) ++len;
return len;
}
public:
ListNode* reverseEvenLengthGroups(ListNode* head) {
ListNode dummy, *tail = &dummy;
for (int i = 1, total = getLength(head); head; ++i) {
int len = min(total, i); // the length of the current section
ListNode *newTail = NULL; // The new tail if we need to reverse the current section
for (int j = 0; j < len && head; ++j) {
auto p = head;
head = head->next;
if (len % 2) { // direct append
tail->next = p;
tail = p;
} else { // append in reverse order
if (newTail == NULL) newTail = p;
p->next = tail->next;
tail->next = p;
}
}
if (newTail) tail = newTail;
total -= len;
}
tail->next = NULL;
return dummy.next;
}
};
https://leetcode.com/problems/reverse-nodes-in-even-length-groups/discuss/1576917/C%2B%2B-Two-Pass