算法优解（7）-相加链表

来自左神书中的一道题，在左神核心代码的基础上，添加了链表的构建和输出操作，将这道题完善成了一个小Demo，和各位共勉。 

题目：两个单链表生成相加链表：

假设链表中的每一节点的值都在0-9之间，于是我们可以使用整个链表作为一个整数。 
例如：链表9->3->7（代表整数937）和链表6->3（代表整数63）相加后得到链表1->0->0->0（代表整数1000）。

思路：利用栈结构求解，关注进位信息。

核心算法：

	public static Node addList(Node head1,Node head2){
		Stack<Integer> s1 = new Stack<Integer>();
		Stack<Integer> s2 = new Stack<Integer>();
		
		while(head1 != null){
			s1.push(head1.value);
			head1 = head1.next;
		}
		while(head2 != null){
			s2.push(head2.value);
			head2 = head2.next;
		}
		
		int ca = 0;
		int n1 = 0;
		int n2 = 0;
		int n = 0;
		
		Node node = null;
		Node pre = null;
		
		while(!s1.isEmpty() || !s2.isEmpty()){
			n1 = s1.isEmpty()?0:s1.pop();
			n2 = s2.isEmpty()?0:s2.pop();
			n = n1 + n2 + ca;
			pre = node;
			node = new Node(n%10);
			node.next = pre;
			ca = n / 10;
		}
		if(ca == 1){
			pre = node;
			node = new Node(1);
			node.next = pre;
		}
		return node;
	}

完整源代码：

import java.util.Stack;

public class AddedListTest {

	/**
	 * 权兴权意-2016.11.14
	 * 题目：两个单链表生成相加链表：
	 * 假设链表中的每一节点的值都在0-9之间，于是我们可以使用整个链表作为一个整数。 
	 * 例如：链表9->3->7（代表整数937）和链表6->3（代表整数63）相加后得到链表1->0->0->0（代表整数1000）。
	 */
	public static void main(String[] args) {
		//构建链表0-4
		Node head1 = new Node(0);
		Node temp1 = head1;
		for(int i = 1;i < 5;i++){
			temp1.next = new Node(i);
			temp1 = temp1.next;
		}
		printList(head1);
		//构建链表5-9
		Node head2 = new Node(5);
		Node temp2 = head2;
		for(int i = 6;i < 10;i++){
			temp2.next = new Node(i);
			temp2 = temp2.next;
		}
		printList(head2);
		
		Node added = addList(head1, head2);
		printList(added);		
		
	}
	
	public static Node addList(Node head1,Node head2){
		Stack<Integer> s1 = new Stack<Integer>();
		Stack<Integer> s2 = new Stack<Integer>();
		
		while(head1 != null){
			s1.push(head1.value);
			head1 = head1.next;
		}
		while(head2 != null){
			s2.push(head2.value);
			head2 = head2.next;
		}
		
		int ca = 0;
		int n1 = 0;
		int n2 = 0;
		int n = 0;
		
		Node node = null;
		Node pre = null;
		
		while(!s1.isEmpty() || !s2.isEmpty()){
			n1 = s1.isEmpty()?0:s1.pop();
			n2 = s2.isEmpty()?0:s2.pop();
			n = n1 + n2 + ca;
			pre = node;
			node = new Node(n%10);
			node.next = pre;
			ca = n / 10;
		}
		if(ca == 1){
			pre = node;
			node = new Node(1);
			node.next = pre;
		}
		return node;
	}
	
	//打印链表
	public static void printList(Node head){
		Node temp = head;
		while(temp != null){
			System.out.print(temp.value + "->");
			temp = temp.next;
		}
		System.out.println();
	}

}