求两个单链表的和

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题目

两个单链表（singly linked list），每一个节点里面一个0-9的数字，输入就相当于两个大数了。然后返回这两个数的和（一个新list）。这两个输入的list 长度相等。要求是：1. 不用递归。2. 要求算法在最好的情况下，只遍历两个list一次，最差的情况下两遍。

实现代码：

public class AddSinglyLinkList {public static void main(String[] args) {int[] a = {1,1,1};int[] b = {1,1,1};int[] a2 = {1, 2, 3};int[] b2 = {1, 2, 8};int[] a3 = {1, 2, 3, 4, 5};int[] b3 = {1, 7, 6, 5, 9};Node head1 = makeLinkedList(a3);Node head2 = makeLinkedList(b3);System.out.println("****************");printNodeList(head1);printNodeList(head2);System.out.println("****************");Node r = addSinglyList(head1, head2);printNodeList(r);}static Node addSinglyList(Node h1, Node h2) {Node p,q;p = h1;q = h1.next;h2 = h2.next;while(q != null) {q.data = q.data + h2.data;//System.out.println(q.data);if(q.data > 9) {q.data = q.data%10;p.data ++;while(p.next != q){p = p.next;p.data = 0;} }else if(q.data != 9) { p = p.next;}q = q.next;h2 = h2.next;}return h1;}static void printNodeList(Node h) {Node n = h.next;while(n != null) {System.out.print(n.data + " ");n = n.next;}System.out.println();}static Node makeLinkedList(int[] a){Node head = new Node(0, null);Node p = head;for(int i: a) {Node temp = new Node(i, null);p.next = temp;p = temp;}return head;}}class Node{int data;Node next;Node(int data, Node next) {this.data = data;this.next = next;}}

参考：http://hawstein.com/posts/add-singly-linked-list.html