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.
描述：
给定两个非空链表，分别表示两个非负整数。整数的组成数字按照逆序排列，并且链表的每个节点表示一个数字。将两个整数相加，并以链表的形式返回相加和。
假设两个整数均不包含前导0(除了整数0本身）。
例子：
输入：(2 -> 4 -> 3) + (5 -> 6 -> 4) 输出：7 -> 0 -> 8 解释： 342 + 465 = 807
遍历两个数字，并逐节点相加。以变量carrier表示进位。
/**
* 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 *p1 = l1;
ListNode *p2 = l2;
int carrier = 0;
ListNode *l3 = NULL;
ListNode *p = NULL;
while(p1 != NULL || p2 != NULL) {
int val1 = p1 == NULL ? 0 : p1->val;
int val2 = p2 == NULL ? 0 : p2->val;
p1 = p1 == NULL ? NULL : p1->next;
p2 = p2 == NULL ? NULL : p2->next;
int val = val1 + val2 + carrier;
carrier = val >= 10 ? 1 : 0;
val = val >= 10 ? val - 10 : val;
ListNode *node = new ListNode(val);
if(l3 == NULL) {
l3 = p = node;
} else {
p->next = node;
p = node;
}
}
if(carrier) {
ListNode *node = new ListNode(carrier);
p->next = node;
}
return l3;
}
};
时间复杂度为O(max(n1, n2)), 空间复杂度为O(max(n1, n2) + 1)。