LeetCode | 86. 分隔链表

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

Code: 巧设双指针

/**
 * 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) {
        //设立头结点可以方便在后面连接链表
        ListNode* before_head = new ListNode(0); ListNode* before = before_head;
        ListNode* after_head  = new ListNode(0); ListNode* after  = after_head;
	
        while (head != nullptr) {
            if (head->val < x) {
                before->next = head;
                before = before->next;
            }
            else {
                after->next = head;
                after = after->next;
            }
            head = head->next;
        }
        after->next = nullptr;
        before->next = after_head->next;
        return before_head->next;
    }
};