标签：链表
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
这道题目主要考察对创建链表的运用而已，注意：表头是个位：
/**
* 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* head = NULL;
ListNode* cur = NULL;
int ad = 0;
while(l1 && l2)
{
if(NULL ==head)
{
head = cur = new ListNode( (l1->val + l2->val + ad) % 10 );
}
else
{
cur->next = new ListNode( (l1->val + l2->val + ad) % 10 );
cur = cur->next;
}
ad = (l1->val + l2->val + ad ) / 10;
l1 = l1->next;
l2 = l2->next;
}
while(l1)
{
cur->next = new ListNode( (l1->val + ad) % 10 );
ad = (l1->val + ad) / 10;
cur = cur->next;
l1 = l1->next;
}
while(l2)
{
cur->next = new ListNode( (l2->val + ad) % 10 );
ad = (l2->val + ad) / 10;
cur = cur->next;
l2 = l2->next;
}
if (ad > 0)
{
cur->next = new ListNode(ad);
}
return head;
}
};再附上大神的代码：
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
ListNode *ret = NULL;
ListNode **pCur = &ret;
int nxt = 0;
while (l1 && l2) {
*pCur = new ListNode((l1->val + l2->val + nxt) % 10);
nxt = (l1->val + l2->val + nxt) / 10;
pCur = &((*pCur)->next);
l1 = l1->next;
l2 = l2->next;
}
while (l1 != NULL) {
*pCur = new ListNode((l1->val + nxt) % 10);
nxt = (l1->val + nxt) / 10;
pCur = &((*pCur)->next);
l1 = l1->next;
}
while (l2 != NULL) {
*pCur = new ListNode((l2->val + nxt) % 10);
nxt = (l2->val + nxt) / 10;
pCur = &((*pCur)->next);
l2 = l2->next;
}
if (nxt > 0) {
*pCur = new ListNode(nxt);
}
return ret;
}
};
标签：链表
原文地址：http://blog.csdn.net/u012771236/article/details/38781727
