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Copy pathproblem2_Add_Two_Numbers.cpp
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89 lines (89 loc) · 1.79 KB
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/**仍然是又臭又长，自己再看一遍都有问题 Runtime:26ms
**1、计算出较短链表的长度（后面的for循环以此为循环次数）
**2、先计算出第一位的结果，创建head，为结果链表的第一个链表
**3、循环相加
**Editorial Solution十分简洁！！然后还体会到三目运算符的强大
**/
/**
* 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 len1 = 0, len2 = 0, len = 0, carry = 0;
ListNode *lp1 = l1, *lp2 = l2, *head = NULL, *lp = NULL;
while (lp1->next)
{
len1++;
lp1 = lp1->next;
}
while (lp2->next)
{
len2++;
lp2 = lp2->next;
}
len = (len1>len2) ? len2 : len1;
lp1 = l1;
lp2 = l2;
head = new ListNode(carry);
lp = head;
head->val = lp1->val + lp2->val;
lp1 = lp1->next;
lp2 = lp2->next;
for (int i = 0; i < len; i++)
{
if (lp->val >= 10)
{
lp->val -= 10;
carry = 1;
}
lp->next = new ListNode(carry);
lp = lp->next;
lp->val += lp1->val + lp2->val;
carry = 0;
lp1 = lp1->next;
lp2 = lp2->next;
}
if (!lp1 && ! lp2)
{
if(lp->val >= 10)
{
lp->next = new ListNode(1);
lp->val -= 10;
lp = lp->next;
}
}
else
{
//取较长的链表的剩余部分
ListNode *last = (lp1 == NULL) ? lp2 : lp1;
while (last)
{
if (lp->val >= 10)
{
lp->val -= 10;
carry = 1;
}
lp->next = new ListNode(carry);
lp = lp->next;
lp->val += last->val;
carry = 0;
last = last->next;
}
}
//最后一位，检查是否大于9
if (lp->val >= 10)
{
lp->val -= 10;
lp->next = new ListNode(1);
lp = lp->next;
}
lp->next = NULL;
return head;
}
};