链表加法的两种实现

第一种是高位在链表尾部
a = 7->1->2
b = 4->3
result = 1->5->2

没啥说的，注意进位，注意链表长度不一定一样

#include<stdio.h>
#include<stdlib.h>
 
/* Linked list node */
struct node
{
    int data;
    struct node* next;
};
 
/* Function to create a new node with given data */
struct node *newNode(int data)
{
    struct node *new_node = (struct node *) malloc(sizeof(struct node));
    new_node->data = data;
    new_node->next = NULL;
    return new_node;
}
 
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(struct node** head_ref, int new_data)
{
    /* allocate node */
    struct node* new_node = newNode(new_data);
 
    /* link the old list off the new node */
    new_node->next = (*head_ref);
 
    /* move the head to point to the new node */
    (*head_ref)    = new_node;
}
 
/* Adds contents of two linked lists and return the head node of resultant list */
struct node* addTwoLists (struct node* first, struct node* second)
{
    struct node* res = NULL; // res is head node of the resultant list
    struct node *temp, *prev = NULL;
    int carry = 0, sum;
 
    while (first != NULL || second != NULL) //while both lists exist
    {
        // Calculate value of next digit in resultant list. 
        // The next digit is sum of following things
        // (i)  Carry
        // (ii) Next digit of first list (if there is a next digit)
        // (ii) Next digit of second list (if there is a next digit)
        sum = carry + (first? first->data: 0) + (second? second->data: 0);
 
        // update carry for next calulation
        carry = (sum >= 10)? 1 : 0;
 
        // update sum if it is greater than 10
        sum = sum % 10;
 
        // Create a new node with sum as data
        temp = newNode(sum);
 
        // if this is the first node then set it as head of the resultant list
        if(res == NULL)
            res = temp;
        else // If this is not the first node then connect it to the rest.
            prev->next = temp;
 
        // Set prev for next insertion
        prev  = temp;
 
        // Move first and second pointers to next nodes
        if (first) first = first->next;
        if (second) second = second->next;
    }
 
    if (carry > 0)
      temp->next = newNode(carry);
 
    // return head of the resultant list
    return res;
}
 
// A utility function to print a linked list
void printList(struct node *node)
{
    while(node != NULL)
    {
        printf("%d ", node->data);
        node = node->next;
    }
    printf("
");
}
 
/* Drier program to test above function */
int main(void)
{
    struct node* res = NULL;
    struct node* first = NULL;
    struct node* second = NULL;
 
    // create first list 7->5->9->4->6
    push(&first, 6);
    push(&first, 4);
    push(&first, 9);
    push(&first, 5);
    push(&first, 7);
    printf("First List is ");
    printList(first);
 
    // create second list 8->4
    push(&second, 4);
    push(&second, 8);
    printf("Second List is ");
    printList(second);
 
    // Add the two lists and see result
    res = addTwoLists(first, second);
    printf("Resultant list is ");
    printList(res);
 
   return 0;
}

第二种是高位在链表头部

a = 2 -> 1 -> 7 
b = 3 -> 4
result = 2 -> 5 -> 1

这个为了处理进位，需要用递归的思想

// A recursive program to add two linked lists

#include <stdio.h>
#include <stdlib.h>

// A linked List Node
struct node
{
    int data;
    struct node* next;
};

typedef struct node node;

/* A utility function to insert a node at the beginning of linked list */
void push(struct node** head_ref, int new_data)
{
    /* allocate node */
    struct node* new_node = (struct node*) malloc(sizeof(struct node));

    /* put in the data  */
    new_node->data  = new_data;

    /* link the old list off the new node */
    new_node->next = (*head_ref);

    /* move the head to point to the new node */
    (*head_ref)    = new_node;
}

/* A utility function to print linked list */
void printList(struct node *node)
{
    while (node != NULL)
    {
        printf("%d  ", node->data);
        node = node->next;
    }
    printf("
");
}

// A utility function to swap two pointers
void swapPointer( node** a, node** b )
{
    node* t = *a;
    *a = *b;
    *b = t;
}

/* A utility function to get size of linked list */
int getSize(struct node *node)
{
    int size = 0;
    while (node != NULL)
    {
        node = node->next;
        size++;
    }
    return size;
}

// Adds two linked lists of same size represented by head1 and head2 and returns
// head of the resultant linked list. Carry is propagated while returning from
// the recursion
node* addSameSize(node* head1, node* head2, int* carry)
{
    // Since the function assumes linked lists are of same size,
    // check any of the two head pointers
    if (head1 == NULL)
        return NULL;

    int sum;

    // Allocate memory for sum node of current two nodes
    node* result = (node *)malloc(sizeof(node));

    // Recursively add remaining nodes and get the carry
    result->next = addSameSize(head1->next, head2->next, carry);

    // add digits of current nodes and propagated carry
    sum = head1->data + head2->data + *carry;
    *carry = sum / 10;
    sum = sum % 10;

    // Assigne the sum to current node of resultant list
    result->data = sum;

    return result;
}

// This function is called after the smaller list is added to the bigger
// lists's sublist of same size.  Once the right sublist is added, the carry
// must be added toe left side of larger list to get the final result.
void addCarryToRemaining(node* head1, node* cur, int* carry, node** result)
{
    int sum;

    // If diff. number of nodes are not traversed, add carry
    if (head1 != cur)
    {
        addCarryToRemaining(head1->next, cur, carry, result);

        sum = head1->data + *carry;
        *carry = sum/10;
        sum %= 10;

        // add this node to the front of the result
        push(result, sum);
    }
}

// The main function that adds two linked lists represented by head1 and head2.
// The sum of two lists is stored in a list referred by result
void addList(node* head1, node* head2, node** result)
{
    node *cur;

    // first list is empty
    if (head1 == NULL)
    {
        *result = head2;
        return;
    }

    // second list is empty
    else if (head2 == NULL)
    {
        *result = head1;
        return;
    }

    int size1 = getSize(head1);
    int size2 = getSize(head2) ;

    int carry = 0;

    // Add same size lists
    if (size1 == size2)
        *result = addSameSize(head1, head2, &carry);

    else
    {
        int diff = abs(size1 - size2);

        // First list should always be larger than second list.
        // If not, swap pointers
        if (size1 < size2)
            swapPointer(&head1, &head2);

        // move diff. number of nodes in first list
        for (cur = head1; diff--; cur = cur->next);

        // get addition of same size lists
        *result = addSameSize(cur, head2, &carry);

        // get addition of remaining first list and carry
        addCarryToRemaining(head1, cur, &carry, result);
    }

    // if some carry is still there, add a new node to the fron of
    // the result list. e.g. 999 and 87
    if (carry)
        push(result, carry);
}

// Driver program to test above functions  
int main()
{
    node *head1 = NULL, *head2 = NULL, *result = NULL;

    int arr1[] = {9, 9, 9};
    int arr2[] = {1, 8};

    int size1 = sizeof(arr1) / sizeof(arr1[0]);
    int size2 = sizeof(arr2) / sizeof(arr2[0]);

    // Create first list as 9->9->9
    int i;
    for (i = size1-1; i >= 0; --i)
        push(&head1, arr1[i]);

    // Create second list as 1->8
    for (i = size2-1; i >= 0; --i)
        push(&head2, arr2[i]);

    addList(head1, head2, &result);

    printList(result);
    getchar();
    return 0;
}