合并排序链表

本文介绍了合并排序链表的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我最近刷牙的一些基础知识，发现合并排序链表是一个pretty的很好的挑战。如果你有一个很好的实现则显示它在这里下车。

I was recently brushing up on some fundamentals and found merge sorting a linked list to be a pretty good challenge. If you have a good implementation then show it off here.

推荐答案

不知道为什么它应该是很大的挑战，因为它是在这里说，这里是Java中的一个简单的实现与任何聪明的技巧。

Wonder why it should be big challenge as it is stated here, here is a straightforward implementation in Java with out any "clever tricks".

//The main function public Node merge_sort(Node head) { if(head == null || head.next == null) { return head; } Node middle = getMiddle(head); //get the middle of the list Node sHalf = middle.next; middle.next = null; //split the list into two halfs return merge(merge_sort(head),merge_sort(sHalf)); //recurse on that } //Merge subroutine to merge two sorted lists public Node merge(Node a, Node b) { Node dummyHead, curr; dummyHead = new Node(); curr = dummyHead; while(a !=null && b!= null) { if(a.info <= b.info) { curr.next = a; a = a.next; } else { curr.next = b; b = b.next; } curr = curr.next; } curr.next = (a == null) ? b : a; return dummyHead.next; } //Finding the middle element of the list for splitting public Node getMiddle(Node head) { if(head == null) { return head; } Node slow, fast; slow = fast = head; while(fast.next != null && fast.next.next != null) { slow = slow.next; fast = fast.next.next; } return slow; }

一些更多的解释在这里 - www.dontforgettothink/2011/11/23/merge-sort-of-linked-list