2. Add Two Numbers
Medium
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.
熟悉链表使用,注意头节点(虚结点)的好处。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode ln(-1); //这个结点是虚结点
ListNode* ptr = &ln;
int sum = 0;int carry = 0;
while(l1||l2||carry)
{
if(l1){
sum+=l1->val;
l1=l1->next;
}
if(l2){
sum+=l2->val;
l2=l2->next;
}
if(carry){
sum+=carry;
carry=0;
}
carry=sum/10;
ptr->next = new ListNode(sum%10);
ptr = ptr->next;
sum = 0;
}
ptr->next = NULL;//最后一个结点置为NULL
ptr = ln.next;//最后通过虚结点返回结果
return ptr;
}
};
博客围绕两非空链表表示非负整数相加展开,数字以逆序存储在链表节点,每个节点含一位数,需将两数相加并以链表形式返回,还提醒熟悉链表使用及头节点好处。
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