Leetcode: Partition List

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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.

Analysis: Linked List 慣用套路，Runner Technique(Two Pointers), 一些技巧就是：設定head的前置假節點prev，兩個pointer：current和runner都指到這個prev，然後進行判斷總是判斷 current.next 或者 runner.next. 這樣做按照我多次做類似題的經驗來說，是最方便省事不容易出錯的。這道題一次過。思路就是current 和 runner 一直移動直到找到 current.next >= x 為止，這裡就是後面小於x的元素將要插入的位置，current便停在這裡，指示這個位置，runner繼續往後面尋找，把每一個小於x的元素都插入到current.next 的位置。

 1 /** 2  * Definition for singly-linked list. 3  * public class ListNode { 4  *     int val; 5  *     ListNode next; 6  *     ListNode(int x) { 7  *         val = x; 8  *         next = null; 9  *     }10  * }11  */12 public class Solution {13     public ListNode partition(ListNode head, int x) {14         ListNode prev = new ListNode(-1);15         prev.next = head;16         ListNode current = prev;17         ListNode runner = prev;18         while (current.next != null && current.next.val < x) {19             current = current.next;20             runner = runner.next;21         }22         while (runner.next != null) {23             if (runner.next.val < x) {24                 ListNode temp = runner.next;25                 runner.next = runner.next.next;26                 temp.next = current.next;27                 current.next = temp;28                 current = current.next;29             }30             else runner = runner.next;31         }32         return prev.next;33     }34 }