Leetcode: Added number represented by linked list
You are given two non-empty linked lists representing two non-negative integers. 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.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Solution
The approach is simple. Loop through each pair from two linked list, add and check if carray
/**
* 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) {
ListNode head(0), *tail, *p;
int carray = 0;
tail = &head;
// loop through each list element
while(l1 || l2){
p = new ListNode(0);
// a+b+carray
if(l1){
p->val += l1->val;
l1 = l1->next;
}
if(l2){
p->val += l2->val;
l2 = l2->next;
}
p->val += carray;
carray = 0;
// if sum > 9 than carray
if(p->val > 9){
carray = p->val / 10;
p->val %= 10;
}
tail->next = p;
tail = tail->next;
}
// check if new sum long than l1 and l2
if(carray){
p = new ListNode(carray);
tail->next = p;
tail = tail->next;
}
// return
return head.next;
}
};Check out the Add Two Numbers.
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