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.
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思路：
用两个链表分别连接小于x和大于x的结点，然后将两个链表合并。
code:
/**
* 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 dummy1(0),dummy2(0); // 两个伪头结点
ListNode *p1 = &dummy1,*p2 = &dummy2; // 两个移动指针
while(head){
if(head->val < x){
p1->next = head;
p1 = p1->next;
}else{
p2->next = head;
p2 = p2->next;
}
head = head->next;
}
p2->next = NULL;
p1->next = dummy2.next;
return dummy1.next;
}
};原文：http://blog.csdn.net/itismelzp/article/details/51167271