算法与数据结构面试题(10)-颠倒链表

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题目

用一种算法来颠倒一个链接表的顺序。现在在不用递归式的情况下做一遍。

解题思路

1.先用递归颠倒

2.尝试不用递归颠倒

代码

1.递归式

public class Problem8 {public LinkedListNode invert(LinkedListNode node) {if (node == null) {throw new NullPointerException();}LinkedListNode endNode = null;LinkedListNode nextNode = node.getNextNode();if (nextNode != null) {node.setNextNode(null);endNode = invert(nextNode);nextNode.setNextNode(node);} else {endNode = node;}return endNode;}public LinkedListNode parseLinkedList(int[] data) {LinkedListNode lastNode = null;for (int i = data.length - 1; i >= 0; i--) {LinkedListNode currentNode = new LinkedListNode();currentNode.setValue(data[i]);currentNode.setNextNode(lastNode);lastNode = currentNode;}return lastNode;}public static void main(String[] args) {int[] data = { 1, 2, 3, 4, 5, 6, 7 };Problem8 problem8 = new Problem8();LinkedListNode rootNode = problem8.parseLinkedList(data);LinkedListNode endNode = problem8.invert(rootNode);problem8.printlnLinkedArray(endNode);System.out.println("Done");}public void printlnLinkedArray(LinkedListNode node){if(node == null) return;System.out.println("" + node.getValue());printlnLinkedArray(node.getNextNode());}}

输出

7654321Done

2.非递归式

public class Problem8_2 {public LinkedListNode invert(LinkedListNode node, int length) {LinkedListNode[] nodes = new LinkedListNode[length];// 先断链for (int i = 0; i < length; i++) {if (node != null) {nodes[i] = node;node = node.getNextNode();}}// 再指针反向for (int i = length - 1; i >= 0; i--) {if (i == 0) {nodes[i].setNextNode(null);} else {nodes[i].setNextNode(nodes[i - 1]);}}return nodes[length - 1];}public LinkedListNode parseLinkedList(int[] data) {LinkedListNode lastNode = null;for (int i = data.length - 1; i >= 0; i--) {LinkedListNode currentNode = new LinkedListNode();currentNode.setValue(data[i]);currentNode.setNextNode(lastNode);lastNode = currentNode;}return lastNode;}public static void main(String[] args) {int[] data = { 1, 2, 3, 4, 5, 6, 7 };Problem8_2 problem8 = new Problem8_2();LinkedListNode rootNode = problem8.parseLinkedList(data);LinkedListNode endNode = problem8.invert(rootNode, data.length);problem8.printlnLinkedArray(endNode);System.out.println("Done");}public void printlnLinkedArray(LinkedListNode node) {if (node == null)return;System.out.println("" + node.getValue());printlnLinkedArray(node.getNextNode());}}

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