[LeetCode]Partition List

题目

Number: 86
Difficulty: Medium
Tags: Linked List, Two Pointers

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.

题解

把链表分成两部分，小于x和大于等于x，保持原来元素的相对位置。

代码

在原链表上移动交换结点：

ListNode* partition(ListNode* head, int x) {    if(!head)        return head;    ListNode *p = head, *q = p, *pre = NULL, *qpre = NULL;    while(q){        while(p && p->val < x){            pre = p;            p = p->next;        }        q = p;        while(q && q->val >= x){            qpre = q;            q = q->next;        }        if(q == NULL)            return head;        if(pre == NULL){            qpre->next = q->next;            q->next = p;            head = q;            pre = head;        }         else{            qpre->next = q->next;            pre->next = q;            q->next = p;            pre = pre->next;        }    }    return head;}

生成两个链表，然后再合并两个链表。

ListNode* partition(ListNode* head, int x) {    ListNode node1(0), node2(0);    ListNode *p = &node1, *q = &node2;    while(head){        if(head->val < x){            p->next = head;            p = p->next;        }        else{            q->next = head;            q = q->next;        }        head = head->next;    }    q->next = NULL;    p->next = node2.next;    return node1.next;}