java单链表反转
java单链表反转
今天做leetcode,遇到了单链表反转。研究了半天还搞的不是太懂,先做个笔记吧
参考:http://blog.****.net/guyuealian/article/details/51119499 https://www.cnblogs.com/hiver/p/7008112.html
class Node { public String value; public Node next; public Node(String value) { this.value = value; } }
public class javaTestLink { public static void main(String[] args) { Node A = new Node("A"); Node B = new Node("B"); Node C = new Node("C"); Node D = new Node("D"); Node E = new Node("E"); Node F = new Node("F"); A.next = B; B.next = C; C.next = D; D.next = E; E.next = F; print(A); // 转换之前打印的字符串:A->B->C->D->E->F /* 采用递归,从后往前 */ Node tmpNode = execute2(A); print(tmpNode); // 转化后的字符串:F->E->D->C->B->A /* 采用非递归,从前往后 */ // Node tmp = execute2(A); // print(tmp); //转化后的字符串:F->E->D->C->B->A } private static Node execute(Node node) { // node看作是前一结点,node.next是当前结点,prev是反转后新链表的头结点 Node prev = null; if (node == null || node.next == null) { prev = node; // 若为空链或者当前结点在尾结点,则直接还回 } else { Node tmp = execute(node.next); // 先反转后续节点head.getNext() node.next.next = node; // 将当前结点的指针域指向前一结点 node.next = null; // 前一结点的指针域令为null; prev = tmp; } return prev; } private static Node execute2(Node node) { if (node == null) return node; Node pre = node; // 上一节点 Node cur = node.next;// 当前结点 Node tmp;// 临时节点,用于保存当前结点的指针域(即下一节点) while (cur != null) { // 当前结点为null,说明位于尾结点 tmp = cur.next; cur.next = pre; // 反转指针域的指向 // 指针往下移动 pre = cur; cur = tmp; } // 最后将原链表的头节点的指针域置为null,还回新链表的头结点,即原链表的尾结点 node.next = null; return pre; } private static void print(Node node) { while (node != null) { System.out.print(node.value); node = node.next; if (node != null) { System.out.print("->"); } else { System.out.println(); } } } }
(二)实现反转的方法:
posted on 2017-12-21 12:37 Michael2397 阅读(...) 评论(...) 编辑 收藏