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.

解法一：

直接插入法：

/** * 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) {        if(!head || !head->next) return head;        ListNode* dummy = new ListNode(-1);        dummy->next = head;        ListNode* pre = dummy;        ListNode* cur;        while(pre->next && pre->next->val < x){            pre = pre->next;        }        ListNode* head2 = pre;        pre = head2->next;             while(pre&&pre->next){            cur = pre->next;            if(cur->val < x){                pre->next = cur->next;  //先删除需要插入的节点                cur->next = head2->next;                head2->next =cur;                head2= head2->next;            }            else pre = pre->next;        }        return dummy->next;    }};

解法二：

把原链表拆分成两部分，再合起来即可

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;}