版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/a19990412/article/details/88423629
题目描述
给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。
如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。
您可以假设除了数字 0 之外,这两个数都不会以 0 开头。
输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出:7 -> 0 -> 8
原因:342 + 465 = 807
题解
- 其实就是考验对于cpp指针的理解而已啦。
- 注意下细节就好了
/**
* 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 tmp = 0, flag = 0;
ListNode * head = NULL, ** curr = NULL;
while(l1 != NULL && l2 != NULL){
tmp = l1->val + l2->val + flag;
if (head == NULL) {
head = new ListNode(tmp % 10);
curr = &head;
} else {
(*curr) = new ListNode(tmp % 10);
}
flag = tmp / 10;
curr = &((*curr)->next);
l1 = l1 -> next;
l2 = l2 -> next;
}
while(l1 != NULL) {
tmp = l1->val + flag;
*curr = new ListNode(tmp % 10);
flag = tmp / 10;
if (head == NULL) {
head = new ListNode(tmp % 10);
curr = &head;
} else {
(*curr) = new ListNode(tmp % 10);
}
curr = &((*curr)->next);
l1 = l1->next;
}
while(l2 != NULL) {
tmp = l2->val + flag;
*curr = new ListNode(tmp % 10);
flag = tmp / 10;
if (head == NULL) {
head = new ListNode(tmp % 10);
curr = &head;
} else {
(*curr) = new ListNode(tmp % 10);
}
curr = &((*curr)->next);
l2 = l2->next;
}
while (flag != 0) {
tmp = flag;
*curr = new ListNode(tmp % 10);
flag = tmp / 10;
if (head == NULL) {
head = new ListNode(tmp % 10);
curr = &head;
} else {
(*curr) = new ListNode(tmp % 10);
}
curr = &((*curr)->next);
}
return head;
}
};
看到了大佬的一份用java写的程序,看起来冗余更少,更精致。我改写成cpp学习下
/**
* 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 tmp = 0, flag = 0, x, y;
ListNode * head = NULL, ** curr = NULL;
while(l1 != NULL || l2 != NULL || flag){
x = (l1 == NULL) ? 0: l1 -> val;
y = (l2 == NULL) ? 0: l2 -> val;
tmp = x + y + flag;
if (head == NULL) {
head = new ListNode(tmp % 10);
curr = &head;
} else {
(*curr) = new ListNode(tmp % 10);
}
flag = tmp / 10;
curr = &((*curr)->next);
l1 = (l1 == NULL) ? NULL: l1 -> next;
l2 = (l2 == NULL) ? NULL: l2 -> next;
}
return head;
}
};