标签：
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1‘s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8.
SOLUTION:
很简单，就是对应加法，重新见一个linkedlist，然后还要考虑进位，要一个进位count，位数加法还要加上count。
代码：
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { /** * @param l1: the first list * @param l2: the second list * @return: the sum list of l1 and l2 */ public ListNode addLists(ListNode l1, ListNode l2) { if (l1 == null){ return l2; } if (l2 == null){ return l1; } int count = 0; // int sum = 0; ListNode dummy = new ListNode(0); ListNode head = dummy; while (l1 != null && l2 != null){ head.next = new ListNode((l1.val + l2.val + count) % 10); count =(l1.val + l2.val + count) / 10; l1 = l1.next; l2 = l2.next; head = head.next; } while (l1 != null){ head.next = new ListNode((l1.val + count) % 10); count = (l1.val + count) / 10; // head = head.next; l1 = l1.next; head = head.next; } while (l2 != null){ head.next = new ListNode((l2.val + count) % 10); count = (l2.val + count) / 10; // head = head.next; l2 = l2.next; head = head.next; } if (count != 0){ head.next = new ListNode(count); } return dummy.next; } }
FOLLOW UP：
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in forward order, such that the 1‘s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given 6->1->7 + 2->9->5. That is, 617 + 295.
Return 9->1->2. That is, 912.
SOLUTION：
其实就是正向的操作，向正常的加法那样，不过我们既然会倒着的，正着肯定也会啊！我们就先reverse一下，把这些都倒过来，然后再用第一题方法加上，然后再把答案reverse回来，就OK啦！
代码：
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { /** * @param l1: the first list * @param l2: the second list * @return: the sum list of l1 and l2 */ public ListNode addLists2(ListNode l1, ListNode l2) { if (l1 == null){ return l2; } if (l2 == null){ return l1; } l1 = reverse(l1); l2 = reverse(l2); ListNode head = add(l1, l2); return reverse(head); } private ListNode reverse(ListNode head){ if (head == null){ return null; } ListNode pre = null; while (head != null){ ListNode temp = head.next; head.next = pre; pre = head; head = temp; } return pre; } private ListNode add(ListNode l1, ListNode l2){ if (l1 == null){ return l2; } if (l2 == null){ return l1; } int sum = 0; int count = 0; ListNode dummy = new ListNode(0); ListNode curr = dummy; while (l1 != null && l2 != null){ sum = l1.val + l2.val + count; curr.next = new ListNode(sum % 10); count = sum / 10; curr = curr.next; l1 = l1.next; l2 = l2.next; } while (l1 != null){ sum = l1.val + count; curr.next = new ListNode(sum % 10); count = sum / 10; curr = curr.next; l1 = l1.next; } while (l2 != null){ sum = l2.val + count; curr.next = new ListNode(sum % 10); count = sum / 10; curr = curr.next; l2 = l2.next; } if (count != 0){ curr.next = new ListNode(count); } return dummy.next; } }
[Lintcode] Add Two Numbers I && II
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原文地址：http://www.cnblogs.com/tritritri/p/4971259.html
