LeetCode: Partition List

LeetCode: 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.

地址：https://oj.leetcode.com/problems/partition-list/

算法：首先，找到第一个大于等于x的节点，记其前趋节点为pre，则在继续往后遍历，若发现比x小的值，则把该节点从链表中移出来，插入到pre节点后面。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) {
        if(!head)   return NULL;
        ListNode *p = head;
        ListNode *pre = NULL;
        while(p && p->val < x){
            pre = p;
            p = p->next;
        }
        if(!p)
            return head;
        ListNode *q = NULL;
        while(p->next){
            if(p->next->val < x){
                q = p->next;
                p->next = q->next;
                if(pre){
                    q->next = pre->next;
                    pre->next = q;
                    pre = q;
                }else{
                    q->next = head;
                    head = q;
                    pre = q;
                }
            }else{
                p = p->next;
            }
        }
        return head;
    }
};