问题：

 

 

package com.nxz.blog.otherTest;public class TestLeetCode {    public static void main(String[] args) {        TestLeetCode t = new TestLeetCode();        ListNode l1 = new ListNode(2);        ListNode l2 = new ListNode(4);        ListNode l3 = new ListNode(3);        ListNode r1 = new ListNode(5);        ListNode r2 = new ListNode(6);        ListNode r3 = new ListNode(4);        l1.next = l2;        l2.next = l3;        r1.next = r2;        r2.next = r3;        ListNode listNode = t.addTwoNumbers(l1, r1);        do {            System.out.println(listNode.val);        } while ((listNode = listNode.next )!= null);    }    /**     * 循环 l1 和 l2 两个listnode，当l1 或 l2 不为null的时候，将两个数相加，大于10时进行处理（保存进一位），否则视为0和另一个值相加     *     * @param l1     * @param l2     * @return     */    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    　　　　　//哑结点，最终返回的节点就是该节点的下一个节点        ListNode dummyHead = new ListNode(-1);        //暂存变量        ListNode p = l1, q = l2, curr = dummyHead;        //大于10时的暂存值        int carry = 0;        //当p或q不为null是，将值相加，此时需要处理，其中一个为null的情况        while (p != null || q != null) {            //只要p或q为nul时，将val值默认为0，和另一个值相加            int x = p != null ? p.val : 0;            int y = q != null ? q.val : 0;            int sum = carry + x + y;            //重新设置进一变量            carry = sum / 10;            curr.next = new ListNode(sum % 10);            //重新设置当前节点，p，q节点，以便再一次循环            curr = curr.next;            if (p != null) {                p = p.next;            }            if (q != null) {                q = q.next;            }        }        //最后处理最高为的carry的值        if (carry > 0) {            curr.next = new ListNode(carry);        }        return dummyHead.next;    }    public static class ListNode {        int val;        ListNode next;        ListNode(int x) {            val = x;        }    }}

 

 

 进阶----->>>>

两数相加：节点为正序的情况

package com.nxz.blog.otherTest;import java.util.Stack;public class TestLeetCode {    public static void main(String[] args) {        TestLeetCode t = new TestLeetCode();        ListNode l1 = new ListNode(2);        ListNode l2 = new ListNode(4);        ListNode l3 = new ListNode(3);        ListNode r1 = new ListNode(5);        ListNode r2 = new ListNode(6);        ListNode r3 = new ListNode(4);        l1.next = l2;        l2.next = l3;        r1.next = r2;        r2.next = r3;        ListNode listNode = t.addTwoNumbers(l1, r1);        do {            System.out.println(listNode.val);        } while ((listNode = listNode.next) != null);    }    /**     * 利用栈这种数据结构（先进后出），这样就可以将正序的节点转换为倒叙的节点了，这样依次从两个栈中取出值（顺序就是个、十、百。。。）     * 这样就转换为之前的那种方式了     *     * @param l1     * @param l2     * @return     */    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        Stack
     
       stack1 = new Stack<>();        Stack
      
        stack2 = new Stack<>();        ListNode dummyHead = new ListNode(-1);        ListNode p = l1, q = l2, head = dummyHead;        while (p != null) {            stack1.add(p);            p = p.next;        }        while (q != null) {            stack2.add(q);            q = q.next;        }        p = stack1.pop();        q = stack2.pop();        int carry = 0;        while (p != null || q != null) {            int x = p != null ? p.val : 0;            int y = q != null ? q.val : 0;            int sum = carry + x + y;            carry = sum / 10;            ListNode last = head.next;            head.next = new ListNode(sum % 10);            head.next.next = last;            p = stack1.isEmpty() ? null : stack1.pop();            q = stack2.isEmpty() ? null : stack2.pop();        }        if (carry > 0) {            ListNode last = head.next;            head.next = new ListNode(carry);            head.next.next = last;        }        return dummyHead.next;    }    public static class ListNode {        int val;        ListNode next;        ListNode(int x) {            val = x;        }    }}