Leetcode-86. Partition List

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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.

这个题目一开始理解错了，原来是先找一个大于等于x的数，然后把后面所有小于的移到前面就可以了。Your runtime beats 5.10% of java submissions.

public class Solution {    public ListNode partition(ListNode head, int x) {        if(head == null) return head;        if(head.next == null) return head;        boolean isExits = false;        ListNode headPoint = new ListNode(0);        headPoint.next = head;        ListNode searchNode = head;        ListNode bigNode = null, preNode = headPoint;        searchNode = headPoint;        while(searchNode != null){            if(searchNode.next != null && searchNode.next.val >= x && bigNode == null) {                preNode = searchNode; bigNode = searchNode.next;                isExits = true; break;            }            searchNode = searchNode.next;        }            if(isExits){            ListNode preSearchNode = preNode;            searchNode = bigNode;            while(searchNode != null){                if(searchNode.val < x){                    preSearchNode.next = searchNode.next;                    preNode.next = searchNode;                    searchNode.next = bigNode;                    preNode = preNode.next;                    searchNode = preSearchNode.next;                }else{                    preSearchNode = searchNode;                    searchNode = searchNode.next;                }            }        }        return headPoint.next;    }}

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