《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 = 0;
        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 / 10;
            val %= 10;
            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 / 10;
            val %= 10;
            tail->next = new ListNode(val);
            tail = tail->next;
            p1 = p1->next;
        }
        while (p2 != nullptr) {
            val = p2->val + carry;
            carry = val / 10;
            val %= 10;
            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) == 2 && n1 > 0 && n2 > 0) {
        // create two linked lists
        ptr = head1 = nullptr;
        for (i = 0; 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 = 0; 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 0;
}