Algorithms—86.Partition List

思路：按照要求，把节点分为2个list，递归。然后再合并。

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode partition(ListNode head, int x) {    List<ListNode> littleList=zuheL(head,new ArrayList<ListNode>(),x);    List<ListNode> bigList=zuheB(head,new ArrayList<ListNode>(),x);    int ll=littleList.size();    int lb=bigList.size();    if (ll==0||lb==0) {    return head;}    for (int i = 0; i < ll-1; i++) {    littleList.get(i).next=littleList.get(i+1);}    littleList.get(ll-1).next=bigList.get(0);    for (int i = 0; i < lb-1; i++) {    bigList.get(i).next=bigList.get(i+1);}    bigList.get(lb-1).next=null;    return littleList.get(0);    }    public List<ListNode> zuheL(ListNode head,List<ListNode> list,int x){    if (head!=null) {    if (head.val<x) {    list.add(head);}list=zuheL(head.next,list,x);}    return list;    }    public List<ListNode> zuheB(ListNode head,List<ListNode> list,int x){    if (head!=null) {    if (head.val>=x) {    list.add(head);}list=zuheB(head.next,list,x);}    return list;    }}

耗时：312ms，中上游。