LeetCode（86） Partition List

题目

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.

分析

给定一个链表与一个目标值x，要求对链表结点按照以下要求重排：

（1）值小于x的结点，按照其原顺序排列与链表头部

（3）其余值不小于x的结点，按照其原顺序链接与尾部

AC代码

class Solution {
public:
    ListNode* partition(ListNode* head, int x) {

        if (!head || !head->next)
            return head;

        ListNode *p = head;
        head = NULL;
        ListNode *r = head;

        //(1)寻找第一个不小于目标值的节点p，前n个小于x的节点逐个连接到head，保存尾结点r
        while (p && p->val < x)
        {
            if (!head)
            {
                head = p;
                r = head;
            }
            else{
                r->next = p;
                //保存新链表尾结点
                r = r->next;
            }
            p = p->next;
            r->next = NULL;

        }//while

        //如果此时p为空，已经没有其余节点，返回新链表head
        if (!p)
            return head;

        //（2）p节点大于x，从下一个结点开始遍历，将小于x的结点连接到head中，原链表删除该结点
        ListNode *pre = p , *q = p->next;
        while (q)
        {
            if (q->val < x)
            {

                if (!head)
                {
                    head = q;
                    r = head;
                }
                else{
                    r->next = q;
                    //保存新链表尾结点
                    r = r->next;
                }

                pre->next = q->next;
                q = q->next;

                r->next = NULL;
            }//if
            else{
                pre = q;
                q = q->next;
            }//else
        }//while

        //如果此时，原链表还剩余结点，直接连接到r后面即可
        if (p)
        {
            if (!head)
                return p;
            else
            r->next = p;
        }

        return head;
    }
};

GitHub测试程序源码