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LeetCode: Partition List [086]

2024-07-06 17:09:44   224人阅读

【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.



【题意】

给定一个链表和一个数值x，将链表中的值按x进行划分，小于x的在前，大于等于x的在后。两部分中节点的之间的相对位置与在原链表中时相同


【思路】

 先将链表按x分裂成两个链表，一个链表中的值小于x, 另一个链表中的值大于等于x
 然后在将两个链表链接起来。


【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) {
        if(head==NULL)return head;
        
        ListNode*headLess=NULL;
        ListNode*tailLess=NULL;
        ListNode*headGreater=NULL;
        ListNode*tailGreater=NULL;
        ListNode*pointer=head;
        while(pointer){
            if(pointer->val<x){
                if(tailLess==NULL)headLess=pointer;
                else tailLess->next=pointer;
                tailLess=pointer;
                pointer=pointer->next;
                tailLess->next=NULL;
            }
            else{
                if(tailGreater==NULL)headGreater=pointer;
                else tailGreater->next=pointer;
                tailGreater=pointer;
                pointer=pointer->next;
                tailGreater->next=NULL;
            }
        }
        //合并两个链表
        if(tailLess){
            head=headLess;
            tailLess->next=headGreater;
        }
        else head=headGreater;
        
        return head;
    }
};


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