Partition List 链表划分区间@LeetCode

简单的partition，用两个链表，3个指针。

package Level4;import Utility.ListNode;/** * 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 and x = 3,return 1->2->2->4->3->5. * */public class S86 {public static void main(String[] args) {}public ListNode partition(ListNode head, int x) {        ListNode l1 = new ListNode(-1);// 构建dummy链表头，l1存放所有比x小的节点        ListNode l2 = new ListNode(-1);        ListNode cur = head;        ListNode l1cur = l1;        ListNode l2cur = l2;                while(cur != null){        if(cur.val < x){        l1cur.next = new ListNode(cur.val);        l1cur = l1cur.next;        }else{        l2cur.next = new ListNode(cur.val);        l2cur = l2cur.next;        }        cur = cur.next;        }                l1cur.next = l2.next;// 把两个链表合并起来        return l1.next;    }}

/** * 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) {        ListNode leftHead = new ListNode(-1);        ListNode rightHead = new ListNode(-1);                ListNode left = leftHead, right = rightHead;        while(head != null) {            if(head.val < x) {                left.next = head;                left = left.next;            } else {                right.next = head;                right = right.next;            }            head = head.next;        }        right.next = null;      // Notice we need to cut off the ring        left.next = rightHead.next;        return leftHead.next;    }}