标签：log list sum nod inter mos amp number new
You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.
Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.
Example:
Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 8 -> 0 -> 7
思路一：
题目说了代表两个int型的数，第一个思路是我直接将这个list转成int型的正整数， 然后两个int型相加，为了防止溢出，用long存储sum；
再将sum转换成list；
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
unsigned int getNum(ListNode *pnode) {
unsigned int result = 0;
while(pnode != NULL) {
result = result*10 + pnode->val;
pnode = pnode->next;
}
return result;
}
ListNode* transNumToList(long num) {
ListNode *cur = new ListNode(num%10);
num = num/10;
while (num > 0) {
int res = num%10;
num /= 10;
ListNode *pnew = new ListNode(res);
pnew->next = cur;
cur = pnew;
}
return cur;
}
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
unsigned int num1 = getNum(l1);
unsigned int num2 = getNum(l2);
cout << num1 << endl;
cout << num2 << endl;
unsigned long sum = num1+num2;
return transNumToList(sum);
}
};
Leetcode刷题总结: 445. Add Two Numbers II
标签：log list sum nod inter mos amp number new
原文地址：http://www.cnblogs.com/syjbupt/p/7497929.html
