Code & Fun

题目是Add Two Numbers II:

题目描述:

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

求解思路:

这道题最简练的方法就是用栈了做了,然后还可以先计算出两个链表的长度,再生成链表。

思路都比较简单,emmm,一开始以为做成递归的会比较简单,后来发现不仅时间复杂度还是代码的复杂度都挺高的,做成迭代的会好很多。

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
/**
* 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) {

stack<int> s1, s2;
int carry = 0, v1, v2, v;
ListNode *cur;
ListNode *pre = nullptr;

while(l1 || l2) {
if (l1) { s1.push(l1->val); l1 = l1->next; }
if (l2) { s2.push(l2->val); l2 = l2->next; }
}


while(!s1.empty() || !s2.empty() || carry) {
v1 = v2 = 0;
if (!s1.empty()) { v1 = s1.top(); s1.pop(); }
if (!s2.empty()) { v2 = s2.top(); s2.pop(); }

v = v1 + v2 + carry;
cur = new ListNode(v % 10);
cur->next = pre;
pre = cur;
carry = v /10;
}
return pre;
}
};

给出一个dicuss中的方法,很巧妙,用的是翻转链表(题目说不能修改原来的链表,所以这里是翻转output):

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
int n1 = 0, n2 = 0, carry = 0;
ListNode *curr1 = l1, *curr2 = l2, *res = NULL;
while( curr1 ){ curr1=curr1->next; n1++; }
while( curr2 ){ curr2=curr2->next; n2++; }
curr1 = l1; curr2 = l2;
while( n1 > 0 && n2 > 0){
int sum = 0;
if( n1 >= n2 ){ sum += curr1->val; curr1=curr1->next; n1--;}
if( n2 > n1 ){ sum += curr2->val; curr2=curr2->next; n2--;}
res = addToFront( sum, res );
}
curr1 = res; res = NULL;
while( curr1 ){
curr1->val += carry; carry = curr1->val/10;
res = addToFront( curr1->val%10, res );
curr2 = curr1;
curr1 = curr1->next;
delete curr2;
}
if( carry ) res = addToFront( 1, res );
return res;
}

ListNode* addToFront( int val, ListNode* head ){
ListNode* temp = new ListNode(val);
temp->next = head;
return temp;
}

本文首发于Code & Fun