Cracking the coding interview--Q2.4

Cracking the coding interview--Q2.4

December 16, 2012
作者：Hawstein

出处：http://hawstein.com/posts/2.4.html

声明：本文采用以下协议进行授权： 自由转载-非商用-非衍生-保持署名|Creative Commons BY-NC-ND 3.0 ，转载请注明作者及出处。

题目

原文：

You have two numbers represented by a linked list, where each node
contains a single digit. The digits are stored in reverse order,
such that the 1’s digit is at the head of the list. Write a function
that adds the two numbers and returns the sum as a linked list.

EXAMPLE

Input: (3 –> 1 –> 5), (5 –> 9 –> 2)

Output: 8 –> 0 –> 8

译文：

你有两个由单链表表示的数。每个结点代表其中的一位数字。数字的存储是逆序的，
也就是说个位位于链表的表头。写一函数使这两个数相加并返回结果，结果也由链表表示。

例子：(3 –> 1 –> 5), (5 –> 9 –> 2)

输入：8 –> 0 –> 8

解答

这道题目并不难，需要注意的有：1.链表为空。2.有进位。3.链表长度不一样。
代码如下：

package cha2;

public class B024 {

static class Node {
int value;
Node next;
public Node() {
value = -1;
next = null;
}
public Node(int value, Node next) {
this.value = value;
this.next = next;
}
}

public static Node add(Node r1, Node r2) {
int carry = 0;
Node sum = new Node(0,null);
if (r1 == null)
return r2;
if (r2 == null)
return r1;
Node tmp = sum;
Node end = new Node(0, null);
while (r1!=null && r2!=null) {
tmp.value = ( r1.value + r2.value + carry ) % 10;
carry = ( r1.value + r2.value + carry ) / 10;
Node tn = new Node(0, null);
tmp.next = tn;
r1 = r1.next;
r2 = r2.next;
end = tmp;
tmp = tn;
}
if (r1==null && r2 == null) {
if (carry == 0)
tmp = null;
else
tmp.value = 1;
}
else {
end.next = (r1 != null) ? r1 : r2;
tmp = end.next;
while (carry == 1) {
tmp.value = tmp.value + carry;
carry = tmp.value  / 10;
tmp.value = tmp.value % 10;
if (tmp.next != null)
tmp = tmp.next;
else {
Node tn = new Node(0, null);
tmp.next = tn;
tmp = tn;
}
}
}
return sum;
}

public static void main(String[] args) {
Node t2 = new Node(9,null);
Node t1 = new Node(9,t2);
Node r1 = new Node(6, t1);
Node r2 = new Node(7, null);
Node sum = add(r1, r2);
while (sum!=null) {
System.out.print(sum.value);
sum = sum.next;
}
}

}