86.Partition List

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：此题的任务是划分链表，比x小的链表节点都出现在比x大或者等于的节点前面，而且还要保持原来的相对位置不要发生变化。我们新建两个链表了l1,l2，遍历原来的链表 ，把值小于x的节点都加入l1,把值大于或者等于x的节点都加入l2，最后将l2连接至l1的后面，所得的链表就是我们所求的结果了。

/**
 * 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 *head1,*head2;
       head1=head2=NULL;
       ListNode *cur1,*cur2;
       cur1=cur2=NULL;
       while(head){
        ListNode *cur=head;
        head=head->next;
        cur->next=NULL;
        if(cur->val<x){
            if(!head1){
                head1=cur;
                cur1=head1;
            }
            else{
                cur1->next=cur;   
                cur1=cur1->next;
            }
            
        }else{
            if(!head2){
                head2=cur;
                cur2=head2;
            }else{
                cur2->next=cur;
                cur2=cur2->next;
            }
        }
       }
        if(!cur1)
            return head2;
        cur1->next=head2;
        return head1;
        
        
        
        
    }
};

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