10.两个链表相加求和

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我是靠谱客的博主缓慢绿草，这篇文章主要介绍10.两个链表相加求和，现在分享给大家，希望可以做个参考。

题目：假设链表中每一个节点的值都在 0 - 9 之间，那么链表整体就可以代表一个整数。

给定两个这种链表，请生成代表两个整数相加值的结果链表。

思路：首先是逆序相加；故这里用到了两个栈，分别存放两个链表中的值。两数相加会涉及到进位问题，所以考虑了进位变量。每次计算完毕都创建一个新的结点，然后连接，最后得到新的链表。

复制代码

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        Stack<Integer> stack1 = new Stack();
        Stack<Integer> stack2 = new Stack();
        while(head1 != null){
            stack1.push(head1.val);
            head1 = head1.next;
        }
        while(head2 != null){
            stack2.push(head2.val);
            head2 = head2.next;
        }
        int n = 0;
        int n1 = 0;
        int n2 = 0;
        int ca = 0;
        ListNode pre = null;
        ListNode node = null;
        while(!stack1.isEmpty() || !stack2.isEmpty()){
            n1 = stack1.isEmpty()?0:stack1.pop();
            n2 = stack2.isEmpty()?0:stack2.pop();
            n = n1 + n2 +ca;
            pre = node;
            node = new ListNode(n%10);
            node.next = pre;
            ca = n/10;
            
        }
        if(ca == 1){
            pre = node;
            node = new ListNode(1);
            node.next = pre;
        }
        return node;
    }
}

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import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 * }
 */

public class Solution {
    /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        Stack<Integer> stack1 = new Stack();
        Stack<Integer> stack2 = new Stack();
        while(head1 != null){
            stack1.push(head1.val);
            head1 = head1.next;
        }
        while(head2 != null){
            stack2.push(head2.val);
            head2 = head2.next;
        }
        int n = 0;
        int n1 = 0;
        int n2 = 0;
        int ca = 0;
        ListNode pre = null;
        ListNode node = null;
        while(!stack1.isEmpty() || !stack2.isEmpty()){
            n1 = stack1.isEmpty()?0:stack1.pop();
            n2 = stack2.isEmpty()?0:stack2.pop();
            n = n1 + n2 +ca;
            pre = node;
            node = new ListNode(n%10);
            node.next = pre;
            ca = n/10;
            
        }
        if(ca == 1){
            pre = node;
            node = new ListNode(1);
            node.next = pre;
        }
        return node;
    }
}