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原题网址(中):https://leetcode-cn.com/problems/add-two-numbers/
原题网址(英):https://leetcode.com/problems/add-two-numbers/
题目:
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.
给出两个非空的链表用来表示两个非负的整数。其中,它们各自的位数是按照逆序的方式存储的,并且它们的每个节点只能存储一位数字。
如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外,这两个数都不会以 0 开头。
示例:
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
思路:
从低位到高位运算,每次求和若大于10,则向高位直接进1.
最开始想的是采用递归直接调用自身函数,但是直接在第一个样例直接被pass了,因为超过时间限制,耗时103ms…只好采用直接while循环来做。
代码:
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0);
ListNode pNode = head;
while (l1 != null || l2 != null) {
int num;//当前位求和
if (l1 == null) {
num = l2.val;
l2 = l2.next;
} else if (l2 == null) {
num = l1.val;
l1 = l1.next;
} else {
num = l1.val + l2.val;
l1 = l1.next;
l2 = l2.next;
}
if (num >= 10) {//如果当前位求和大于0,则高位(下节点)直接加一。
num = num - 10;
if (l1 != null) {
l1.val += 1;
} else {
l1 = new ListNode(1);
l1.next = null;
}
}
pNode.next = new ListNode(num);
pNode = pNode.next;
}
return head.next;
}
}