題目要求
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.
例子:
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.
思路:
每一位相加,當前位推入鏈表,記錄進位數值,每一次的當前位都等於兩個當前位數字相加再加上進位數字。
要點:
- 複習一下c++中地址的使用、類的定義,鏈表是最簡單的常用自定義數據類型之一了。
- 因爲是單鏈表,沒有
prev
,所以既要定義起始node的地址,也要記錄node的變量名,node的地址在迭代過程中不斷在變,爲了最後還能找到起始點。 - 比較特殊的情況就是兩個數字的位數不同,這種情況下可以在該位補0。
代碼:
/**
* 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 StartFrom(0);
ListNode *p = &StartFrom;
int carry=0;
while (l1 || l2 || carry)
{
int term1 = l1 ? l1->val : 0;
int term2 = l2 ? l2->val : 0;
int sum = term1 + term2 + carry;
carry = sum / 10;
int digits = sum % 10;
p->next = new ListNode(digits);
p = p->next;
l1 = l1? l1->next : 0;
l2 = l2? l2->next : 0;
}
return StartFrom.next;
}
};