LEETCODE专题
Add Two Numbers
首先先上题目:
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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
下面是一些个人理解和细节把握:
链表的每个节点其实都可以相加然后往下走到下一个节点,但是要把握好前提条件:
- 两个链表要保持同一个深度
- 每一个节点相加后的进位要存储至结果链表的下一个节点
这些都是可以做到的,下面有代码,在这里不再赘述
2.在1[1]的基础上提出问题:倘若两个链表不等长怎么办?
这时候就加一个条件判断就能够解决问题
3.最后倘若两个输入链表已经结束然而进位为1,则要在结果链表的尾端新建一个结果节点,将进位存进去
下面直接上代码:
/**
* 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) {
bool first = true; // firstly enter the while block
int carry = 0; // carry = 1 or 0
ListNode * sum = new ListNode(0); // store the sum of l1 and l2
ListNode * head_of_sum = sum; // return the head of ListNode * sum
while (l1 != NULL || l2 != NULL) {
int temp_sum = carry;
if (first == false) {
sum->next = new ListNode(0);
sum = sum->next;
} else {
first = false;
}
if (l1 == NULL) {
temp_sum += l2->val;
} else if (l2 == NULL) {
temp_sum += l1->val;
} else {
temp_sum += l1->val + l2->val;
}
carry = temp_sum / 10;
sum->val = temp_sum % 10;
if (l1 != NULL) {
l1 = l1->next;
}
if (l2 != NULL) {
l2 = l2->next;
}
}
if (carry == 1) {
sum->next = new ListNode(carry);
}
return head_of_sum;
}
};