[Algorithm] Daily practice two and the number of LeetCode02

Title LeetCode02 adding two numbers

We are given two non-empty list is used to represent two non-negative integer. Where their respective bits are stored in reverse order of the way, and they each node can store only one digit.

If we add up these two numbers, it will return a new list and to represent them.

You can assume that in addition to the numbers 0, these two numbers will not begin with 0.

Example:

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

Code

package com.zixin.algorithm;

public class LC02 {

	public static void main(String[] args) {
		ListNode l1 = new ListNode(2);
		ListNode temp1 = new ListNode(4);
		ListNode temp2 = new ListNode(3);
		l1.next = temp1;
		temp1.next = temp2;

		ListNode l2 = new ListNode(5);
		ListNode temp3 = new ListNode(6);
		ListNode temp4 = new ListNode(4);
		l2.next = temp3;
		temp3.next = temp4;
		ListNode node = addTwoNumbers(l1, l2);
		while (node != null) {
			System.out.println(node.val);
			node = node.next;
		}

	}

	/**
	 * 
	 * @param l1
	 * @param l2
	 * @return
	 * @Desc 注意点：1、如何返回结果对应的链表   pre.next
	 *            2、加的结果可能会出现进位
	 *            3、最后也可能会产生进位
	 *            4、两个链表长度不一致的时候，后面要补0
	 */
	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		ListNode pre = new ListNode(0);
		ListNode cur = pre;
		int carry = 0;
		while (l1 != null || l2 != null) {// 判断两个链表是否为null
			int x = l1 == null ? 0 : l1.val;// 如果为null为补0，否则就是取该节点的值
			int y = l2 == null ? 0 : l2.val;
			int sum = x + y + carry;// 默认进位的地方是1

			carry = sum / 10;// 这里是处理是否有进位
			sum = sum % 10;// 这里是处理当前的结果
			cur.next = new ListNode(sum);// 根据结果构建下一个节点

			cur = cur.next;// 当前节点指向下一个节点
			if (l1 != null) {// 不为空 向后走
				l1 = l1.next;
			}
			if (l2 != null) {
				l2 = l2.next;
			}
		}
		if (carry == 1) {// 循环完之后，如果最后一个有进位，则需要单独处理
			cur.next = new ListNode(carry);
		}
		return pre.next;// 返回pre.next的原因是，最开始创建的时候有一个pre节点，从pre的下一个节点开始，才是真正的数据节点的开头
	}

}

class ListNode {
	int val;
	ListNode next;

	ListNode(int x) {
		val = x;
	}
}

Author Information
Name: Zi Xin
qq exchange group: 1027372701