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Leetcode: Partition List

2024-07-09 11:59:58   226人阅读

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 惯用套路，Runner Technique(Two Pointers), 一些技巧就是：设置head的前置假节点prev，两个pointer：current和runner都指到这个prev，然后进行判断总是判断 current.next 或者 runner.next. 这样做按照我多次做类似题的经验来说，是最方便省事不容易出错的。这道题一次过。思路就是current 和 runner 一直移动直到找到 current.next >= x 为止，这里就是后面小于x的元素将要插入的位置，current便停在这里，指示这个位置，runner继续往后面寻找，把每一个小于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 }

 

 


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