partition-list

Posted 2020-07-18 Adding

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,
Given1->4->3->2->5->2and x = 3,
return1->2->2->4->3->5.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode partition(ListNode head, int x) {
           
        if(head==null)
            return null;
        ListNode  smallHead=new ListNode(0);
        ListNode  bigHead=new ListNode(0);
        ListNode small=smallHead;
        ListNode big=bigHead;
        ListNode p=head;
        while(p!=null){
            if(p.val>=x){
                big.next=p;
                big=big.next;
            }else{
                small.next=p;
                small=small.next;
            }
            p=p.next;
        }
        small.next=bigHead.next;
                big.next=null;//必须把后面断掉，因为原来的节点还指着后面。我这里犯错误了，没有断掉，导致报错
        return smallHead.next;
        
    }
    
}