Partition List

lc86

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.

思路一是两个node，然后合并 感觉面试会这个思路即可

    public ListNode partition(ListNode head, int x) {
        if (head == null) {
            return null;
        }

        ListNode leftDummy = new ListNode(0);
        ListNode rightDummy = new ListNode(0);
        ListNode left = leftDummy, right = rightDummy;

        while (head != null) {
            if (head.val < x) {
                left.next = head;
                left = head;
            } else {
                right.next = head;
                right = head;
            }
            head = head.next;
        }

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

思路二：用两个pointer

 public ListNode partition(ListNode head, int x) {  
        if(head == null)  
            return null;  
        ListNode dummy = new ListNode(0);  
        dummy.next = head;  
        ListNode walker = dummy;  
        ListNode runner = dummy;  
        while(runner.next!=null)  
        {  
            if(runner.next.val<x)  
            {  
                if(runner!=walker)  
                {  
                    ListNode next = runner.next.next;  
                    runner.next.next = walker.next;  
                    walker.next = runner.next;  
                    runner.next = next;  
                }  
                else  
                    runner = runner.next;  
                walker = walker.next;  
            }  
            else  
            {  
                runner = runner.next;  
            }  
        }  
        return dummy.next;  

    }

与partition array的方法对笔着来

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        if (head == null) return head;
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode start = dummy;
        head = dummy;
        while(head.next != null) {
            if (head.next.val < x) {
                if (start != head) {
                    ListNode temp = start.next;
                    start.next = head.next;
                    ListNode temp2 = head.next.next;
                    head.next.next = temp;
                    start = start.next;
                    head.next = temp2; 
                } else {
                    head = head.next;
                    start = start.next;
                }

            } else {
                head = head.next;
            }
        }
        return dummy.next;
    }
}