LeetCode算法入門- 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：

主要注意兩個問題：進位問題；鏈表長度不同問題。

/*** 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);//定義另一個ListNode curr，另其等同于newHead，//至于為什么不直接用newHead，因為最后要通過newHead來返回頭部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){//最后還有記得判斷最后一個carry的值有沒有進位curr.next = new ListNode(carry);}//記得返回的是newHead的next而非newHeadreturn newHead.next;} } 創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎勵來咯，堅持創(chuàng)作打卡瓜分現(xiàn)金大獎

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