problem:
You are given two linked lists representing two non-negative numbers. 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. Input: (2 -> 4 -> 3) + (5 -> 6 -> 4) Output: 7 -> 0 -> 8
兩個單鏈表表示的數字相加,再將結果用單鏈表表示出來,考察鏈表的基本操作,注意進位即可
注意:
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
題目給定的兩個鏈表指針l1、l2常理來說指向第一個有效元素,不是指向頭結點的指針。因此,返回的指針也是指向有效結點的指針,不是指向頭結點。
還有:鏈表最好使用帶頭結點的表示方法,方便鏈表操作!!頭結點的數據域無用,隨便初始化,用的是其指針域。
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) {
int flag = 0;
ListNode* tail = new ListNode(0);
ListNode* ptr = tail;
while(l1 != NULL || l2 != NULL){
int val1 = 0;
if(l1 != NULL){
val1 = l1->val;
l1 = l1->next;
}
int val2 = 0;
if(l2 != NULL){
val2 = l2->val;
l2 = l2->next;
}
int tmp = val1 + val2 + flag;
ptr->next = new ListNode(tmp % 10);
flag = tmp / 10;
ptr = ptr->next;
}
if(flag == 1){
ptr->next = new ListNode(1);
}
return tail->next;
}
};