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.

 

方法：扫一遍链表，通过与x比较，将链表分成两个链表，一个链表是<x的节点集，一个链表是>=x的节点集，代码如下：

/**
 * 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) {
        if( !head ) return 0;
        ListNode lnode(0);
        ListNode rnode(0);
        ListNode* lpre = &lnode;    //<x的节点集，保证了节点间的相对顺序
        ListNode* rpre = &rnode;    //>=x的节点集，保证了节点间的相对顺序
        while( head ) {
            if( head->val < x ) {
                lpre->next = head;
                lpre = head;
                head = head->next;
                lpre->next = 0;
            }
            else {
                rpre->next = head;
                rpre = head;
                head = head->next;
                rpre->next = 0;
            }
        }
        lpre->next = rnode.next;
        return lnode.next;
    }
};