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.

思路

本题没有必要在原链表上进行插入删除操作，可以额外建立两个链表，比x小的为一边，反之在另一边

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) {    ListNode node1(0), node2(0);    ListNode *p1 = &node1, *p2 = &node2;    while (head) {        if (head->val < x)            p1 = p1->next = head;        else            p2 = p2->next = head;        head = head->next;    }    p2->next = NULL;    p1->next = node2.next;    return node1.next;    }};