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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("\n"); } /* 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("\n"); } // 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; }
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原文地址:http://www.cnblogs.com/yueyanglou/p/4915413.html