05 August 2008
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
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Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
给两个链表，个位十位百位从左到右存储，两个链表相加返回一个新链表。这道题也是链表的基本操作，创建一个dummy，原先的两个链表挨个结点相加，然后注意下进位就行了。时间O(m+n)，空间创建了一个新的链表O(m)，m是长度较长的那个链表。
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/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
if (l1 == null || l2 == null) {
return l1 == null ? l2 : l1;
}
ListNode dummy = new ListNode(-1);
ListNode current = dummy;
int carry = 0;
while (l1 != null || l2 != null) {
int sum = carry;
if (l1 != null) {
sum += l1.val;
l1 = l1.next;
}
if (l2 != null) {
sum += l2.val;
l2 = l2.next;
}
current.next = new ListNode(sum % 10);
current = current.next;
carry = sum / 10;
}
if (carry != 0) {//循环结束后最后判断下是否还有进位
current.next = new ListNode(carry);
}
return dummy.next;
}
}