题目:
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
代码:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* partition(ListNode* head, int x) {
ListNode* first = new ListNode(0);
ListNode* last = first, *p = head;
first->next = head; bool has = false;
while (head) {
if (head->val >= x) {
p = head;
head = head->next;
has = true;
}
else {
if (has) {
p->next = head->next;
head->next = last->next;
last->next = head;
head = p->next;
last = last->next;
}
else {
p = head;
head = head->next;
last = last->next;
}
}
}
return first->next;
}
};