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.

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题目:

有一个数字链表和单独的值x,重新划分链表,

将链表中的所有小于x的节点放到  大于等于x的节点前面,

小于x的节点的相对顺序不变,大于等于x的节点的相对顺序不变.

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思路:

和奇数偶数排序一样类似,

遍历链表的时候,将节点分组,

难点在于怎么判断开始?怎么判断链表结束?

定义两个假的头节点dummy1,dummy2,利用两个p1和p2分别指向前两个节点

外加一个遍历list的指针curr,

最后将dummy1链表的尾指向dummy2链表的头节点(这里dummy1和dummy2都是节点,不是指针).

 

代码如下:

/**
 * 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==nullptr || head->next==nullptr) return head;
        ListNode *head1,*head2;
        ListNode *p1,*p2,*curr;
        ListNode dummy1(-1);
        ListNode dummy2(-1);
        curr = head;
        p1 = head1 = &dummy1;
        p2 = head2 = &dummy2;

        while(curr){
            if(curr->val < x){
                p1->next = curr;
                p1 = p1->next;
                //p1->next = nullptr;
            }else{
                p2->next = curr;
                p2 = p2->next;
                //p2->next = nullptr;
            }
            curr = curr->next;
        }
        p1->next = nullptr;
        p2->next = nullptr;
        //showList(head1->next);
        //showList(head2->next);

        p1->next = head2->next;
        return head1->next;
    }
};