You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
解答
寒假刷的第一题……
相当于是实现一个全加器阵列……大概是还没有走出数逻的阴影……
当然局部变量要记得初始化，将Carry变量赋值为0，这样链表头节点的相加就可以归入链表中一般节点的相加，将Ans赋值为NULL，为循环中链表的插入做准备，每一位都要加上前一位Carry的值并重新计算Carry，第一位的前一位Carry的值是0，最后一位相加也要重新计算Carry，并且如果两个链表长度不同，程序进行到后面只对一个链表进行计算时，每一位也要加上前一位Carry的值并重新计算Carry，因为可能存在9+999999的情况，其实这样的情况也就相当于是一个全加器单元的一个加数是0，并且最后一位相加结束后Carry不为0，则要再增加一位。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
struct ListNode *Ans, *TempNode, *TailNode;
int Carry;
Carry = ;
Ans = TailNode = NULL;
while(NULL != l1 && NULL != l2){
TempNode = (struct ListNode*)malloc(sizeof(struct ListNode));
TempNode->val = (Carry + l1->val + l2->val) % ;
TempNode->next = NULL;
Carry = (Carry + l1->val + l2->val) / ;
if(NULL == Ans){
Ans = TempNode;
TailNode = TempNode;
}
else{
TailNode->next = TempNode;
TailNode = TailNode->next;
}
l1 = l1->next;
l2 = l2->next;
}
while(NULL != l1){
TempNode = (struct ListNode*)malloc(sizeof(struct ListNode));
TempNode->val = (Carry + l1->val) % ;
TempNode->next = NULL;
Carry = (Carry + l1->val) / ;
TailNode->next = TempNode;
TailNode = TailNode->next;
l1 = l1->next;
}
while(NULL != l2){
TempNode = (struct ListNode*)malloc(sizeof(struct ListNode));
TempNode->val = (Carry + l2->val) % ;
TempNode->next = NULL;
Carry = (Carry + l2->val) / ;
TailNode->next = TempNode;
TailNode = TailNode->next;
l2 = l2->next;
}
!= Carry){
TempNode = (struct ListNode*)malloc(sizeof(struct ListNode));
TempNode->val = Carry;
TempNode->next = NULL;
TailNode->next = TempNode;
}
return Ans;
}