lintcode-221-链表求和 II

221-链表求和 II

思路

code

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
/*
* @param l1: The first list.
* @param l2: The second list.
* @return: the sum list of l1 and l2.
*/
ListNode * addLists2(ListNode * l1, ListNode * l2) {
// write your code here
l1 = reverse(l1);
l2 = reverse(l2);
return reverse(addLists(l1, l2));
}

ListNode *reverse(ListNode *head) {
// write your code here
ListNode *l1 = NULL, *l2 = NULL, *l3 = NULL;

l1 = head;
//  链表没有节点或有一个节点
if (l1 == NULL || l1->next == NULL) {
return l1;
}
l2 = l1->next;
//  链表有2节点
if (l2->next == NULL) {
l2->next = l1;
l1->next = NULL;
return l2;
}
l3 = l2->next;
//  链表有3个以上节点
if (l2->next != NULL) {
while (l2 != l3) {
l2->next = l1;
if (l1 == head)
l1->next = NULL;
l1 = l2;
l2 = l3;

if (l3->next != NULL)
l3 = l3->next;
}
l2->next = l1;
return l2;
}
}

ListNode *addLists(ListNode *l1, ListNode *l2) {
// write your code here
ListNode *head = new ListNode(0);
ListNode *temp = head;
int carry = 0;
while (l1 != NULL && l2 != NULL) {
int sum = l1->val + l2->val + carry;
ListNode *node = new ListNode(sum % 10);
carry = (sum >= 10);
temp->next = node;
l1 = l1->next;
l2 = l2->next;
temp = temp->next;
}
while (l1 != NULL) {
int sum = l1->val + carry;
ListNode *node = new ListNode(sum % 10);
carry = (sum >= 10);
temp->next = node;
l1 = l1->next;
temp = temp->next;
}
while (l2 != NULL) {
int sum = l2->val + carry;
ListNode *node = new ListNode(sum % 10);
carry = (sum >= 10);
temp->next = node;
l2 = l2->next;
temp = temp->next;
}
if (l1 == NULL && l2 == NULL && carry == 1) {
ListNode *node = new ListNode(carry);
temp->next = node;
}
return head->next;
}
};