LeetCode 2_Add Two Numbers
题目描述：
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
题目很清晰（虽然我看了半天），无非链表相加！
注意：
1、链表长度没说一样长。
2、考虑最后一个节点有可能进位，此时需要新建节点。
首先不加思索写出的代码：
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
{
ListNode* l3 = l1;
int tmp1 = 0, tmp = 0;
while(l1->next != NULL && l2->next != NULL)
{
tmp1 = l1->val;
l1->val = (l1->val + l2->val + tmp) % 10;
tmp = (tmp1 + l2->val + tmp) / 10;
l1 = l1->next;
l2 = l2->next;
}
if(l1->next == NULL && l2->next == NULL)
{
tmp1 = l1->val;
l1->val = (l1->val + l2->val + tmp) % 10;
tmp = (tmp1 + l2->val + tmp) / 10;
if(tmp == 1)
l1->next = new ListNode(1);
return l3;
}
if(l1->next != NULL)
{
tmp1 = l1->val;
l1->val = (l1->val + l2->val + tmp) % 10;
tmp = (tmp1 + l2->val + tmp) / 10;
l1 = l1->next;
while(l1->next != NULL)
{
if(l1->val + tmp < 10)
{
l1->val = l1->val + tmp;
return l3;
}
else
{
l1->val = (l1->val + tmp) % 10;
tmp = 1;
l1 = l1->next;
}
}
if(l1->val + tmp >= 10)
{
l1->val = (l1->val + tmp) % 10;
l1->next = new ListNode(1);
return l3;
}
else
{
l1->val = (l1->val + tmp) % 10;
return l3;
}
}
if(l2->next != NULL)
{
tmp1 = l1->val;
l1->val = (l1->val + l2->val + tmp) % 10;
tmp = (tmp1 + l2->val + tmp) / 10;
l1->next = l2->next;
l1 = l1->next;
while(l1->next != NULL)
{
if(l1->val + tmp < 10)
{
l1->val = l1->val + tmp;
return l3;
}
else
{
l1->val = (l1->val + tmp) % 10;
tmp = 1;
l1 = l1->next;
}
}
if(l1->val + tmp >= 10)
{
l1->val = (l1->val + tmp) % 10;
l1->next = new ListNode(1);
return l3;
}
else
{
l1->val = (l1->val + tmp) % 10;
return l3;
}
}
}
这代码，看起来好low，首先代码重复性太大，从代码的编写就可以有大量的优化空间。
最终优化为（其实也没什么优化，只是删掉了一些重复）：
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
{
ListNode* l3 = l1;
int tmp1 = 0, tmp = 0;
while(l1->next != NULL && l2->next != NULL)
{
tmp1 = l1->val;
l1->val = (l1->val + l2->val + tmp) % 10;
tmp = (tmp1 + l2->val + tmp) / 10;
l1 = l1->next;
l2 = l2->next;
}
tmp1 = l1->val;
l1->val = (l1->val + l2->val + tmp) % 10;
tmp = (tmp1 + l2->val + tmp) / 10;
if(l1->next == NULL && l2->next == NULL)
{
if(tmp == 1)
l1->next = new ListNode(1);
return l3;
}
else
{
if(l2->next != NULL)
l1->next = l2->next;
l1 = l1->next;
while(l1->next != NULL)
{
if(l1->val + tmp < 10)
{
l1->val = l1->val + tmp;
return l3;
}
else
{
l1->val = (l1->val + tmp) % 10;
tmp = 1;
l1 = l1->next;
}
}
if(l1->val + tmp >= 10)
{
l1->val = (l1->val + tmp) % 10;
l1->next = new ListNode(1);
return l3;
}
else
{
l1->val = (l1->val + tmp) % 10;
return l3;
}
}</span>
其实有一种更简单的方法是再新建一个链表，不过就是会浪费空间
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
{
int addNum = 0;
ListNode *l3 = new ListNode(0);
ListNode *ptr = l3;
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;
}
ListNode *temp = new ListNode((val1 + val2 + addNum) % 10);
ptr->next = temp;
ptr = temp;
addNum = (val1 + val2 + addNum) / 10;
}
if(addNum == 1)
ptr->next = new ListNode(1);
return l3->next;
}
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时间： 2024-12-18 10:58:58