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.

这道题开始没什么思路，后面看到提示说用双指针，有点懂了

思路

找出第一个大于等于x的位置front和前一个位置behind,behind后移，如果behind.val < x在front后面插入值为x的节点

public class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode headNode = new ListNode(0);
        headNode.next = head;
        
        ListNode front = head;
        ListNode behind = headNode;            //双指针
        
        while(front != null){                //找出第一个大于等于x的位置和前一个位置
            if(front.val >= x)
                break;
            front = front.next;
            behind = behind.next;
        }//while
        if(front == null)
            return head;                    //如果原列表符合规定直接返回头结点
        while(front != null &&front.next != null){            //遇到比x小的插入到behind后面，同时删除后面的节点
            if(front.next.val < x){
                ListNode temp = new ListNode(front.next.val);
                temp.next = behind.next;
                behind.next = temp;
                behind = temp;
                front.next = front.next.next;                //删除节点
                continue;
            }//if
            front = front.next;
        }//while
        return headNode.next;
    }
}

 这是我在leetcode上面的第100题