Description
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思路
- 很简单的问题，主要是注意进位的问题，以及最后特殊情况，即最后有个进位
代码
- list1和list2的长度分别为n,m。算法复杂度为O(max(n, m))
/*
* 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 *res = NULL, *ptr = NULL;
int count = 0, sum = 0;
while(l1 || l2){
sum = 0;
if(l1){
sum += l1->val;
l1 = l1->next;
}
if(l2){
sum += l2->val;
l2 = l2->next;
}
sum += count;
count = sum / 10;
sum %= 10;
ListNode *tmp = new ListNode(sum);
if(!res){
res = tmp;
ptr = tmp;
}
else{
ptr->next = tmp;
ptr = ptr->next;
}
}
//特殊情况，比如9 + 9 = 18，最后的 1
if(count > 0){
ListNode *tmp = new ListNode(count);
ptr->next = tmp;
}
return res;
}
};内容来源于网络如有侵权请私信删除
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