LeetCode--Add two Number II

LeetCode–Add two Number II

LeetCodeAdd two Number II
题目描述

算法解释

代码

题目描述

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

算法解释

这一次的也是加法，直接是7243+564 = 7807；

要求不能将list倒转，也就是要求，从最高位加起。

一眼就想到的办法，直接求出 7243和564，然后为sum构造list。

（我们要有点创造力）

复习了stack，list

坑

一定是先使用top后pop，否则容易犯非空的错

构造一个逆的list。好好学，认真看。

还是先把list倒转，就i不用花费心思处理其他

代码

class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
stack<int> s1;
stack<int> s2;

while(l1!=NULL) {
s1.push(l1->val);
l1 = l1->next;
}
while(l2!=NULL) {
s2.push(l2->val);
l2 = l2->next;
}

ListNode* list = new ListNode(0);
int sum = 0;
while(!s1.empty() || !s2.empty()) {
if (!s1.empty()) {
sum += s1.top();
s1.pop();
}
if (!s2.empty()) {
sum += s2.top();
s2.pop();
}
list->val = sum % 10;
ListNode* head = new ListNode(sum / 10);
head->next = list;
list = head;
sum /= 10;

}
return list->val >0?list:list->next;

}
};