Add Two Numbers
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 contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
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Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
Output: [8,9,9,9,0,0,0,1]
Constraints:
The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.
C++ Solution
考虑到要验证比较麻烦，先把测试架构实现好, 然后再实现主要逻辑。主要思路就是分别遍历链表，一旦有链表到头了，也不能停止，直到分别指向两个链表的指针都为空，且进位也不为1时，才能够停止整个流程。
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// c++
#include<iostream>
#include<vector>
using namespace std;
struct ListNode {
int val;
ListNode *next;
ListNode() : val(0), next(nullptr) {}
ListNode(int x) : val(x), next(nullptr) {}
ListNode(int x, ListNode *next) : val(x), next(next) {}
};
ListNode* constructLists(vector<int>src) {
ListNode* head = nullptr;
ListNode* prev = nullptr;
for (auto val : src) {
ListNode *current = new ListNode(val);
if (prev != nullptr) {
prev->next = current;
} else {
head = current;
}
prev = current;
}
return head;
}
class Solution {
public:
Solution() {}
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode *ptr1 = l1;
ListNode *ptr2 = l2;
ListNode* head = nullptr;
ListNode* prev = nullptr;
int carry = 0;
int sum = 0;
while (ptr1 != nullptr || ptr2 != nullptr || carry != 0) {
sum = (ptr1 != nullptr ? ptr1->val : 0) +
(ptr2 != nullptr ? ptr2->val : 0) + carry;
carry = sum / 10;
int val = sum % 10;
ListNode *current = new ListNode(val);
if (prev != nullptr) {
prev->next = current;
} else {
head = current;
}
prev = current;
if (ptr1 != nullptr) {
ptr1 = ptr1->next;
}
if (ptr2 != nullptr) {
ptr2 = ptr2->next;
}
}
return head;
}
};
void test(vector<int> vec1, vector<int> vec2, vector<int>target) {
ListNode *list1 = constructLists(vec1);
ListNode *list2 = constructLists(vec2);
Solution *ptr = new Solution();
ListNode *ans = ptr->addTwoNumbers(list1, list2);
ListNode *expect = constructLists(target);
int ansSize = 0;
ListNode *sizeAnsPtr = ans;
while (sizeAnsPtr != nullptr) {
ansSize++;
sizeAnsPtr = sizeAnsPtr->next;
}
if (ansSize != target.size()) {
cout<<"size not matched! expect "<<target.size()<< " real size is "<<ansSize<<endl;
return;
}
ListNode *ansPtr = ans;
ListNode *expectPtr = expect;
bool testRet = true;
while (ansPtr != nullptr && expectPtr != nullptr) {
if (ansPtr->val != expectPtr->val) {
testRet = false;
break;
}
ansPtr = ansPtr->next;
expectPtr = expectPtr->next;
}
if (testRet) {
cout<<"test passed!"<<endl;
} else {
cout<<"test failed!"<<endl;
}
}
int main() {
test(vector<int>{2,4,3}, vector<int>{5,6,4}, vector<int>{7,0,8});
test(vector<int>{0}, vector<int>{0}, vector<int>{0});
test(vector<int>{9,9,9,9,9,9,9}, vector<int>{9,9,9,9}, vector<int>{8,9,9,9,0,0,0,1});
}
