LeetCode.2:Add Two Numbers-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_32350719/article/details/88695820

博客围绕LeetCode题目展开，给定两个非空链表表示非负整数，数字逆序存储，每个节点含一位数。需将两数相加并以链表返回。还给出思路，定义两个结点和整型sum，相加元素赋值给第一个结点，大于10则进位，最后返回第二个结点。

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You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

思路，定义两个结点，定义一个整型sum，sum用来相加两个给定的队列的元素，并将值赋给第一个结点，注意如果相加大于10，需要进一（把一保留下来加到下次循环），第二个结点用来保存第二个结点初始时的位置，最后就返回第二个结点即可

package com.RyanWang;
public class AddTwoNumber {
	public static class ListNode {
		int val;
		ListNode next;
		ListNode(int x) {
			val = x;
		}
	}

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		ListNode l1 = new ListNode(2);
		l1.next = new ListNode(4);
		l1.next.next = new ListNode(3);
		ListNode l2 = new ListNode(5);
		l2.next = new ListNode(6);
		l2.next.next = new ListNode(4);
		ListNode l3;
		l3 = addTwoNumbers(l1, l2);
		for (; l3 != null; l3 = l3.next) {
			System.out.println(l3.val);
		}
	}

	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		int sum = 0;
		ListNode p, tmp = new ListNode(0);
		p = tmp;
		while (l1 != null || l2 != null ||sum != 0) {
			if (l1 != null) {
				sum += l1.val;
				l1 = l1.next;
			}
			if (l2 != null) {
				sum += l2.val;
				l2 = l2.next;
			}
			p.next = new ListNode(sum % 10);
			sum /= 10;
			p = p.next;

		}
		return tmp.next;
	}

}