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.
Example:
Input: head = 1->4->3->2->5->2, x = 3 Output: 1->2->2->4->3->5
Related Topics:
Linked List, Two Pointers
// OJ: https://leetcode.com/problems/partition-list/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode ltHead, geHead, *ltTail = <Head, *geTail = &geHead;
while (head) {
auto p = head;
head = head->next;
if (p->val < x) {
ltTail->next = p;
ltTail = p;
} else {
geTail->next = p;
geTail = p;
}
}
ltTail->next = geHead.next;
geTail->next = NULL;
return ltHead.next;
}
};