Add Two Numbers
You are given two linked lists representing two non-negative numbers. 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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
算法思路：
思路1：
将两个因数转换成整型，求出和，再将和转换为链表形式
【注意】这个int类型会溢出，严格来说，long也会溢出，因此是不严谨的。
代码如下：
1 public class Solution { 2 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 3 if(l1 == null) return l2; 4 if(l2 == null) return l1; 5 return num2List( getNum(l1) + getNum(l2)); 6 } 7 private long getNum(ListNode l){ 8 if(l == null) return 0; 9 long num = 0; 10 long length = 0; 11 while( l != null){ 12 num += Math.pow(10,length) * l.val; 13 length++; 14 l = l.next; 15 } 16 return num; 17 } 18 private ListNode num2List(long num){ 19 ListNode hhead = new ListNode(0); 20 ListNode tail = hhead; 21 if(num == 0) return hhead; 22 while(num != 0){ 23 ListNode tem = new ListNode((int)(num % 10)); 24 tail.next = tem; 25 tail = tail.next; 26 num /= 10; 27 } 28 return hhead.next; 29 } 30 }
思路2：
直接在两个list中进行加和操作。模拟加法。
1 public class Solution { 2 public ListNode addTwoNumbers(ListNode l1, ListNode l2) { 3 if(l1 == null) return l2; 4 if(l2 == null) return l1; 5 int l1Length = getLength(l1),l2Length = getLength(l2); 6 if(l1Length > l2Length) return addTwoNumbers(l2, l1); 7 ListNode l2Pointer = l2; 8 ListNode l1Pointer = l1; 9 while(l2Pointer != null){ 10 l2Pointer.val = l2Pointer.val + ((l1Pointer == null) ? 0 : l1Pointer.val); 11 if(l2Pointer.val > 9){ 12 l2Pointer.val -= 10 ; 13 if(l2Pointer.next == null){ 14 ListNode tail = new ListNode(1); 15 l2Pointer.next = tail; 16 break; 17 } 18 l2Pointer.next.val++; 19 } 20 l2Pointer = l2Pointer.next; 21 if(l1Pointer != null)l1Pointer = l1Pointer.next; 22 } 23 return l2; 24 } 25 private int getLength(ListNode list){ 26 int length = 0; 27 while(list != null){ 28 length++; 29 list = list.next; 30 } 31 return length; 32 } 33 }
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原文：http://www.cnblogs.com/huntfor/p/3863299.html
