《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：

  1 // 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.

  2 // Example First List: 5->6->3 // represents number 365 

  3 // Second List: 8->4->2 // represents number 248 

  4 // Resultant list: 3->1->6 // 

  5 // Note :Any Carry forward should also be added as the new node . Any Comments on the code below

  6 #include <cstdio>

  7 using namespace std;

  8 

  9 struct ListNode {

 10     int val;

 11     ListNode *next;

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

 13 };

 14 

 15 class Solution {

 16 public:

 17     ListNode* listAddition(ListNode *head1, ListNode *head2) {

 18         if (head1 == nullptr) {

 19             return head2;

 20         } else if (head2 == nullptr) {

 21             return head1;

 22         }

 23         

 24         int carry = 0;

 25         int val;

 26         ListNode *p1, *p2;

 27         ListNode *head, *tail;

 28         

 29         p1 = head1;

 30         p2 = head2;

 31         head = tail = nullptr;

 32         while (p1 != nullptr && p2 != nullptr) {

 33             val = p1->val + p2->val + carry;

 34             carry = val / 10;

 35             val %= 10;

 36             if (head == nullptr) {

 37                 head = tail = new ListNode(val);

 38             } else {

 39                 tail->next = new ListNode(val);

 40                 tail = tail->next;

 41             }

 42             p1 = p1->next;

 43             p2 = p2->next;

 44         }

 45         while (p1 != nullptr) {

 46             val = p1->val + carry;

 47             carry = val / 10;

 48             val %= 10;

 49             tail->next = new ListNode(val);

 50             tail = tail->next;

 51             p1 = p1->next;

 52         }

 53         while (p2 != nullptr) {

 54             val = p2->val + carry;

 55             carry = val / 10;

 56             val %= 10;

 57             tail->next = new ListNode(val);

 58             tail = tail->next;

 59             p2 = p2->next;

 60         }

 61         if (carry) {

 62             tail->next = new ListNode(carry);

 63             tail = tail->next;

 64         }

 65         

 66         return head;

 67     }

 68 };

 69 

 70 int main()

 71 {

 72     int i;

 73     int n1, n2;

 74     int val;

 75     struct ListNode *head, *head1, *head2, *ptr;

 76     Solution sol;

 77     

 78     while (scanf("%d%d", &n1, &n2) == 2 && n1 > 0 && n2 > 0) {

 79         // create two linked lists

 80         ptr = head1 = nullptr;

 81         for (i = 0; i < n1; ++i) {

 82             scanf("%d", &val);

 83             if (head1 == nullptr) {

 84                 head1 = ptr = new ListNode(val);

 85             } else {

 86                 ptr->next = new ListNode(val);

 87                 ptr = ptr->next;

 88             }

 89         }

 90         ptr = head2 = nullptr;

 91         for (i = 0; i < n2; ++i) {

 92             scanf("%d", &val);

 93             if (head2 == nullptr) {

 94                 head2 = ptr = new ListNode(val);

 95             } else {

 96                 ptr->next = new ListNode(val);

 97                 ptr = ptr->next;

 98             }

 99         }

100         

101         // add up the two lists

102         head = sol.listAddition(head1, head2);

103         

104         // print the list

105         printf("%d", head->val);

106         ptr = head->next;

107         while (ptr != nullptr) {

108             printf("->%d", ptr->val);

109             ptr = ptr->next;

110         }

111         printf("\n");

112         

113         // delete the list

114         while (head != nullptr) {

115             ptr = head->next;

116             delete head;

117             head = ptr;

118         }

119         while (head1 != nullptr) {

120             ptr = head1->next;

121             delete head1;

122             head1 = ptr;

123         }

124         while (head2 != nullptr) {

125             ptr = head2->next;

126             delete head2;

127             head2 = ptr;

128         }

129     }

130     

131     return 0;

132 }