Add Two Numbers(2)

2— Add Two Numbers

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.

Example :

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

C代码:

// struct ListNode {
//   int val;
//   struct ListNode *next;
// };

//在尾部插入一个Node
int appendList(struct ListNode *l,int data){
  struct ListNode *ptr = l;
  while(ptr && ptr->next != NULL){
    ptr = ptr ->next;
  }
  struct ListNode *tmp = (struct ListNode*)malloc(sizeof(struct ListNode));
  if(!tmp){
    return 0;
  }
  tmp -> val = data;
  tmp -> next = NULL;
  ptr -> next = tmp;
  return 1;

}

struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
    struct ListNode *result = (struct ListNode*)malloc(sizeof(struct ListNode));
    result -> next = NULL;
    int flag = 0;     // 进位标志
    struct ListNode *ptr1 = l1 ,*ptr2 = l2;
    int data1,data2,data3,sum;
    while(ptr1||ptr2){       //将[9,9,9]+[1]->[9,9,9]+[1,0,0]
      data1 = (ptr1 == NULL)? 0:ptr1->val;
      data2 = (ptr2 == NULL)? 0:ptr2->val;
      // printf("d1= %d d2=%d\n",data1,data2 );
      sum = (flag == 1)? data1 + data2 + 1 : data1 + data2;

      data3 = sum % 10;
      flag = sum / 10;
      // printf("d1= %d d2=%d d3=%d\n",data1,data2,data3 );
      appendList(result,data3);
      ptr1 = ptr1?ptr1->next:ptr1;
      ptr2 = ptr2?ptr2->next:ptr2;
    }

    if(flag == 1){       // 最后一个进位,如[9,9]+[1] = [0,0,1]
      appendList(result,1);
    }

    return result->next;

}

C++代码:

// struct ListNode {
//   int val;
//   ListNode *next;
//   ListNode(int x) : val(x), next(NULL) {}
// };

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
      int flag = 0;
      ListNode preHead(0),*p = &preHead;
      while(l1 || l2){
        int sum = (l1?l1->val:0) + (l2?l2->val:0) + flag;
        flag = sum / 10;
        p-> next = new ListNode(sum % 10);
        p = p->next;
        l1 = l1?l1->next : l1;
        l2 = l2?l2->next : l2;
      }
      if(flag == 1){
        p->next = new ListNode(1);
      }
      return preHead.next;
    }
};

Complexity Analysis:

Time complexity : O( $n^{2}$).
Space complexity : O(n).

思路:

• 将999+1 转化成999+001, 即[9,9,9]+[1]->[9,9,9]+[1,0,0], 循环次数为最长链表长度.