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->null and x = 3,
return 1->2->2->4->3->5->null.

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Example
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解题要点 :

哑节点的使用; Dummy

/**
 * Definition for ListNode.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int val) {
 *         this.val = val;
 *         this.next = null;
 *     }
 * }
 */ 
public class Solution {
    /**
     * @param head: The first node of linked list.
     * @param x: an integer
     * @return: a ListNode 
     */
    public ListNode partition(ListNode head, int x) {
        // write your code here
        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode node = head;
        ListNode left = leftDummy;
        ListNode right = rightDummy;
        while(node!=null){
          if (node.val<x) {
            left.next=node;
            left=left.next;
          }else{
            right.next =node;
            right = right.next;
          }
          node=node.next;


        }

    right.next=null;
    left.next=rightDummy.next;
    return leftDummy.next;

    }
}