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Leetcode/problem2_Add_Two_Numbers.cpp at master · DJBDJB/Leetcode · GitHub
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problem2_Add_Two_Numbers.cpp
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Leetcode
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problem2_Add_Two_Numbers.cpp
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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;
}
};
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