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:这题不算难。按x的值分成两部分。详细思路和代码例如以下:

/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode partition(ListNode head, int x) {
/**
* 思路是将list按X分成两段
* 小于的连接p
* 大于的连接q
* 最后合并p和q就可以
*/
ListNode p = new ListNode(0);
ListNode pHead = p;
ListNode q = new ListNode(0);
ListNode qHead = q;
//遍历
while(head != null){
if(head.val < x){//<x成一组
p.next = head;
p = p.next;
}else{//>=x成一组
q.next = head;
q = q.next;
}
head = head.next;
}
p.next = qHead.next;
q.next = null;//斩断后面的连接
return pHead.next;
}
}

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