[LintCode] 96 Partition List 解题报告

Description
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.


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



思路
这题比较简单。做两个新的链表less和greater。过一遍现在的链表，把< x的点连到less，其他所有的连到greater。最后把less和greater连起来就行了。



Code

/**
 * 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
        
        if (head == null || head.next == null) {
            return head;
        }
        
        ListNode lessDummy = new ListNode(0);
        ListNode less = lessDummy;
        ListNode greaterDummy = new ListNode(0);
        ListNode greater = greaterDummy;
        
        ListNode curr = head;
        while (curr != null) {
            if (curr.val < x) {
                less.next = curr;
                less = less.next;
            }
            else {
                greater.next = curr;
                greater = greater.next;
            }
            curr = curr.next;
        }
        
        less.next = greaterDummy.next;
        greater.next = null;
        
        return lessDummy.next;
    }
}