LeetCode算法入门- Add Two Numbers-day3
Add Two Numbers
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.
solution：
主要注意两个问题：进位问题；链表长度不同问题。
/*** 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) {ListNode p1 = l1;ListNode p2 = l2;ListNode newHead = new ListNode(0);//定义另一个ListNode curr，另其等同于newHead，//至于为什么不直接用newHead，因为最后要通过newHead来返回头部ListNode curr = newHead;int carry = 0;while(p1 != null || p2 != null){int x = (p1 != null) ? p1.val : 0;int y = (p2 != null) ? p2.val : 0;int sum = carry + x + y;carry = sum / 10;curr.next = new ListNode(sum % 10);curr = curr.next;if(p1 != null)p1 = p1.next;if(p2 != null)p2 = p2.next;}if(carry > 0){//最后还有记得判断最后一个carry的值有没有进位curr.next = new ListNode(carry);}//记得返回的是newHead的next而非newHeadreturn newHead.next;}
}
