[LeetCode]Partition List

题目描述:(链接)

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.

解题思路:

将链表分为两个链表，一个链表上所有节点的值都小于x，另一个链表为大于等于x，然后连接两个链表。

/**
 * 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 left(-1);
        ListNode right(-1);
        
        ListNode *pLeft = &left;
        ListNode *pRight = &right;
        for (ListNode *cur = head; cur != NULL; cur = cur->next) {
            if (cur->val < x) {
                pLeft->next = cur;
                pLeft = cur;
            } else {
                pRight->next = cur;
                pRight = cur;
            }
        }
        
        pLeft->next = right.next;
        pRight->next = NULL;
        
        return left.next;
    }
};