Partition List

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.

For example,
Given 1->4->3->2->5->2 and x = 3,
return 1->2->2->4->3->5.

思路: 分成两个链表，分别保存bigorgreater 和 less than

java代码：

1. public ListNode partition(ListNode head, int x) {
2. if(head == null) return null;
3. ListNode small = null;
4. ListNode smalltail = null;
5. ListNode big = null;
6. ListNode bigtail = null;
7. ListNode p = head;
8. while(p!=null) {
9. if(p.val >= x) {
10. if(big==null) {
11. big = p;
12. bigtail = p;
13. } else {
14. bigtail.next = p;
15. bigtail = p;
16. }
17. } else {
18. if(small==null) {
19. small = p;
20. smalltail = p;
21. } else {
22. smalltail.next = p;
23. smalltail = p;
24. }
25. }
26. p = p.next;
27. }
28. if(bigtail!=null) bigtail.next = null;  //注意条件判断
29. if(smalltail!=null) smalltail.next = big;
30. if(small==null) return big;
31. else return small;
32. }

思路二：只用一条链表，把小的节点插入到前面，后面的保持不变

C++代码：

1. ListNode *partition(ListNode *head, int x) {
2. if(head == NULL) return NULL;
3. ListNode *dummy = new ListNode(-1);
4. dummy -> next = head;
5. ListNode *p=head->next;
6. ListNode *pre = head;
7. ListNode *q=dummy;
8. if(head->val < x) q=head;
9. else q=dummy;
10. while(p!=NULL) {
11. if(p->val<x) {
12. if(p==q->next) {
13. pre=p;
14. p=p->next;
15. q=q->next;
16. } else{
17. pre->next=p->next;
18. p->next=q->next;
19. q->next=p;
20. q=q->next;
21. p=pre->next;
22. }
23. } else {
24. pre=p;
25. p=p->next;
26. }
27. }
28. return dummy->next;
29. }