问题的本质就是两个单链表的合并；
注意点是：
1.在一条链表结束时，是否还需要进位，如果需要进位，就需要将另外一条单链表循序加进位；
2.在两条链表遍历结束时，是否还有进位，如果有就要在单链表的末尾增加节点；
代码
新单链表
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) {
}
};
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == nullptr && l2 == nullptr) {
return nullptr;
}
ListNode* result = new ListNode;
ListNode* pCurrent = result;
ListNode* pPre = nullptr;
int add = 0;
while (l1 != nullptr && l2 != nullptr) {
int sum = l1->val + l2->val + add;
pCurrent->val = sum % 10;
add = sum / 10;
l1 = l1->next;
l2 = l2->next;
pCurrent->next = new ListNode;
pPre = pCurrent;
pCurrent = pCurrent->next;
}
while (l1 != nullptr)
{
int sum = l1->val + add;
pCurrent->val = sum % 10;
add = sum / 10;
l1 = l1->next;
pCurrent->next = new ListNode;
pPre = pCurrent;
pCurrent = pCurrent->next;
}
while (l2 != nullptr)
{
int sum = l2->val + add;
pCurrent->val = sum % 10;
add = sum / 10;
l2 = l2->next;
pCurrent->next = new ListNode;
pPre = pCurrent;
pCurrent = pCurrent->next;
}
if (add != 0)
{
pCurrent->val = add;
}
else
{
delete pPre->next;
pPre->next = nullptr;
}
return result;
}
};
使用现有单链表
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == nullptr && l2 == nullptr) {
return nullptr;
}
ListNode* result = l1;
ListNode* pre = nullptr;
int add = 0;
while (l1 != nullptr && l2 != nullptr){
int sum = l1->val + l2->val + add;
l1->val = sum % 10;
add = sum / 10;
pre = l1;
l1 = l1->next;
l2 = l2->next;
}
if (l2 != nullptr){
pre->next = l2;
while (l2 != nullptr&&add != 0) {
int sum = l2->val + add;
l2->val = sum % 10;
add = sum / 10;
pre = l2;
l2 = l2->next;
}
}
else {
while (l1 != nullptr && add != 0) {
int sum = l1->val + add;
l1->val = sum % 10;
add = sum / 10;
pre = l1;
l1 = l1->next;
}
}
if (add != 0) {
pre->next = new ListNode(add);
}
return result;
}
};
消灭除法
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
if (l1 == nullptr && l2 == nullptr) {
return nullptr;
}
ListNode* result = l1;
ListNode* pre = nullptr;
int add = 0;
while (l1 != nullptr && l2 != nullptr){
int sum = l1->val + l2->val + add;
l1->val = sum % 10;
add = sum > 9;
pre = l1;
l1 = l1->next;
l2 = l2->next;
}
if (l2 != nullptr){
pre->next = l2;
while (l2 != nullptr&&add != 0) {
int sum = l2->val + add;
l2->val = sum % 10;
add = sum > 9;
pre = l2;
l2 = l2->next;
}
}
else {
while (l1 != nullptr && add != 0) {
int sum = l1->val + add;
l1->val = sum % 10;
add = sum > 9;
pre = l1;
l1 = l1->next;
}
}
if (add != 0) {
pre->next = new ListNode(add);
}
return result;
}
};