Add Two Numbers 题解
原创文章，拒绝转载
题目来源：https://leetcode.com/problems/add-two-numbers/description/
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
Explanation: 342 + 465 = 807.
Solution
class Solution {
private:
ListNode* copyList(ListNode* listHeadNode) {
if (listHeadNode == NULL)
return NULL;
ListNode* resHeadNode = new ListNode(listHeadNode -> val);
ListNode* resCurNode = resHeadNode;
ListNode* listCurNode = listHeadNode -> next;
while (listCurNode != NULL) {
resCurNode -> next = new ListNode(listCurNode -> val);
resCurNode = resCurNode -> next;
listCurNode = listCurNode -> next;
}
return resHeadNode;
}
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == NULL)
return copyList(l2);
if (l2 == NULL)
return copyList(l1);
bool isOverflow = false;
int curVal = l1 -> val + l2 -> val;
if (curVal >= 10) {
isOverflow = true;
}
ListNode* resHeadNode = new ListNode(curVal % 10);
ListNode* resPreNode = resHeadNode;
ListNode* l1CurNode = l1 -> next;
ListNode* l2CurNode = l2 -> next;
while (l1CurNode != NULL && l2CurNode != NULL) {
curVal = l1CurNode -> val + l2CurNode -> val;
if (isOverflow)
curVal++;
resPreNode -> next = new ListNode(curVal % 10);
isOverflow = (curVal / 10 > 0);
resPreNode = resPreNode -> next;
l1CurNode = l1CurNode -> next;
l2CurNode = l2CurNode -> next;
}
if (l1CurNode == NULL)
resPreNode -> next = copyList(l2CurNode);
else
resPreNode -> next = copyList(l1CurNode);
if (isOverflow) {
if (resPreNode -> next == NULL) {
resPreNode -> next = new ListNode(1);
} else {
while (true) {
if (resPreNode -> next == NULL) {
resPreNode -> next = new ListNode(1);
break;
} else {
curVal = resPreNode -> next -> val + 1;
if (curVal >= 10) {
resPreNode -> next -> val = curVal % 10;
resPreNode = resPreNode -> next;
} else {
resPreNode -> next -> val = curVal;
break;
}
}
}
}
}
return resHeadNode;
}
};解题描述
这道题可以说解决的是数字相加的问题，但是由于数字的长度不确定，输入的数字可能是现有的内置数据类型无法存储的，所以像题目中给出的方法是，利用链表来记录数字，然后针对链表中的每一个节点的值进行相加。
明确了题目的要求之后，很快就可以清楚，题目最需要解决的问题就是链表边界和相加进位的问题。我的做法算是简单的对特殊情况打补丁。