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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
这题考的概率比较大，虽然比较简单，但决定还是写一遍。
算法比较简单，直接模拟两数相加，需要注意进位及两数长短不一的情况。
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution {10 public:11 ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {12 if( !l1 ) return l2; //l1为空13 if( !l2 ) return l1; //l2为空14 ListNode node(0); //建立一个头结点15 ListNode* pre = &node; //前驱节点16 int carry = 0; //进位值17 while( l1 && l2 ) { //当两链表所指的节点都存在时18 int sum = l1->val + l2->val + carry; //模拟加法19 ListNode* p = new ListNode( sum%10 );20 carry = sum/10;21 pre->next = p;22 pre = p;23 l1 = l1->next;24 l2 = l2->next;25 }26 l1 = l1 ? l1 : l2; //使l1指向不为空的链表27 while( l1 ) { //处理进位有可能导致高位数变化28 int sum = l1->val + carry;29 ListNode* p = new ListNode( sum%10 );30 carry = sum/10;31 pre->next = p;32 pre = p;33 l1 = l1->next;34 }35 if( carry ) { //如果依然还有进位36 ListNode* p = new ListNode(carry);37 pre->next = p;38 }39 return node.next;40 }41 };
Add Two Numbers
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