《Cracking the Coding Interview》——第2章：链表——题目5

2014-03-18 02:32

题目：给定两个由单链表表示的数字，返回它们的和。比如(9->9) + (1->2) = 0->2->1，99 + 21 = 120。

解法：逐位相加，注意处理进位、长度不等。

代码：

 // 2.5 Given two numbers represented by two lists, write a function that returns sum list. The sum list is list representation of addition of two input numbers.
 // Example First List: 5->6->3 // represents number 365
 // Second List: 8->4->2 // represents number 248
 // Resultant list: 3->1->6 //
 // Note :Any Carry forward should also be added as the new node . Any Comments on the code below
 #include <cstdio>
 using namespace std;

 struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x): val(x), next(nullptr) {};
 };

 class Solution {
 public:
     ListNode* listAddition(ListNode *head1, ListNode *head2) {
         if (head1 == nullptr) {
             return head2;
         } else if (head2 == nullptr) {
             return head1;
         }

         int carry = ;
         int val;
         ListNode *p1, *p2;
         ListNode *head, *tail;

         p1 = head1;
         p2 = head2;
         head = tail = nullptr;
         while (p1 != nullptr && p2 != nullptr) {
             val = p1->val + p2->val + carry;
             carry = val / ;
             val %= ;
             if (head == nullptr) {
                 head = tail = new ListNode(val);
             } else {
                 tail->next = new ListNode(val);
                 tail = tail->next;
             }
             p1 = p1->next;
             p2 = p2->next;
         }
         while (p1 != nullptr) {
             val = p1->val + carry;
             carry = val / ;
             val %= ;
             tail->next = new ListNode(val);
             tail = tail->next;
             p1 = p1->next;
         }
         while (p2 != nullptr) {
             val = p2->val + carry;
             carry = val / ;
             val %= ;
             tail->next = new ListNode(val);
             tail = tail->next;
             p2 = p2->next;
         }
         if (carry) {
             tail->next = new ListNode(carry);
             tail = tail->next;
         }

         return head;
     }
 };

 int main()
 {
     int i;
     int n1, n2;
     int val;
     struct ListNode *head, *head1, *head2, *ptr;
     Solution sol;

     while (scanf("%d%d", &n1, &n2) ==  && n1 >  && n2 > ) {
         // create two linked lists
         ptr = head1 = nullptr;
         for (i = ; i < n1; ++i) {
             scanf("%d", &val);
             if (head1 == nullptr) {
                 head1 = ptr = new ListNode(val);
             } else {
                 ptr->next = new ListNode(val);
                 ptr = ptr->next;
             }
         }
         ptr = head2 = nullptr;
         for (i = ; i < n2; ++i) {
             scanf("%d", &val);
             if (head2 == nullptr) {
                 head2 = ptr = new ListNode(val);
             } else {
                 ptr->next = new ListNode(val);
                 ptr = ptr->next;
             }
         }

         // add up the two lists
         head = sol.listAddition(head1, head2);

         // print the list
         printf("%d", head->val);
         ptr = head->next;
         while (ptr != nullptr) {
             printf("->%d", ptr->val);
             ptr = ptr->next;
         }
         printf("\n");

         // delete the list
         while (head != nullptr) {
             ptr = head->next;
             delete head;
             head = ptr;
         }
         while (head1 != nullptr) {
             ptr = head1->next;
             delete head1;
             head1 = ptr;
         }
         while (head2 != nullptr) {
             ptr = head2->next;
             delete head2;
             head2 = ptr;
         }
     }

     return ;
 }