【LeetCode】Partition List ——链表排序问题

【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.

【解析】

题意：给定一个单链表和一个x，把链表中小于x的放到前面，大于等于x的放到后面，每部分元素的原始相对位置不变。

思路：其实很简单，遍历一遍链表，把小于x的都挂到head1后，把大于等于x的都放到head2后，最后再把大于等于的链表挂到小于链表的后面就可以了。

 

/**
 * 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 lessHead(0);
        ListNode greatHead(0);
        ListNode *cur=head,*less=&lessHead,*great=&greatHead;
        while(cur!=NULL){
            ListNode *next=cur->next;
            if(cur->val<x){
                less->next=cur;
                less=less->next;
                less->next=NULL;
            }
            else{
                great->next=cur;
                great=great->next;
                great->next=NULL;
            }
            cur=next;
        }
        less->next=greatHead.next;
        return lessHead.next;
    }
};