題目：

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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.

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鏈表類：

class ListNode{int val;ListNode next;public ListNode(int val) {this.val = val;} }

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實現(xiàn)類：

public static ListNode addTwoNumbers(ListNode l1 , ListNode l2){ListNode l = null;int temp = 0;l = new ListNode(l1.val+l2.val);ListNode q = new ListNode(0);//q指向第一個結(jié)點q = l;while(l1.next!=null && l2.next!=null){if(l.val<10){l.next =new ListNode(l1.next.val+l2.next.val);}else{temp = l.val;l.val = temp%10;l.next = new ListNode(temp/10+l1.next.val+l2.next.val);}l = l.next;l1 = l1.next;l2 = l2.next;}while (l1.next != null && l2.next == null) {if (l.val>=10) {temp = l.val;l.val = temp%10;l.next = new ListNode(temp/10+l1.next.val);}else {l.next = new ListNode(l1.next.val);}l = l.next;l1 = l1.next;}while (l1.next == null && l2.next != null) {if (l.val>=10) {temp = l.val;l.val = temp%10;l.next = new ListNode(temp/10+l2.next.val);}else {l.next = new ListNode(l2.next.val);}l = l.next;l2 = l2.next;}//處理最后一個數(shù)，是否需要進位if (l.val>=10) {temp = l.val;l.val = temp%10;l.next = new ListNode(temp/10);}l = q;return l;}

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注意點：

1，定義指針q指向第一個結(jié)點。獲得所求的結(jié)果后，將鏈表l指向q，即鏈表的第一個節(jié)點。

2，鏈表l1,l2相加時，需考慮進位問題。

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轉(zhuǎn)載于:https://www.cnblogs.com/wkcode/p/8663922.html

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