Add Two Numbers(medium)

最新推荐文章于 2021-07-27 23:56:12 发布

原创最新推荐文章于 2021-07-27 23:56:12 发布 · 418 阅读

CC 4.0 BY-SA版权

本文介绍了一种使用链表实现两个非负整数相加的方法。通过遍历两个链表节点并考虑进位情况，最终返回一个新的链表表示两数之和。

You are given two linked lists representing two non-negative numbers. 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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

我的解答：

class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
  int n = 0;
  ListNode* a = NULL;
  ListNode* current = NULL;
  if ((l1->val + l2->val) >= 10) {
   n = 1;
   a = new ListNode(l1->val + l2->val - 10);
  }
  else {
   a = new ListNode(l1->val + l2->val);
  }
  current = a;
  l1 = l1->next;
  l2 = l2->next;
  while (l1 != NULL || l2 != NULL||n==1) {
   if (l1 == NULL&&l2 == NULL) {
    current->next = new ListNode(1);
    n = 0;
   }
   else if (l1 == NULL) {
    if (l2->val + n >= 10) {
     current->next = new ListNode(l2->val + n - 10);
     current = current->next;
     n = 1;
    }
    else {
     current->next = new ListNode(l2->val + n);
     current = current->next;
     n = 0;
    }
    l2 = l2->next;
   }
   else if (l2 == NULL) {
    if (l1->val + n >= 10) {
     current->next = new ListNode(l1->val + n - 10);
     current = current->next;
     n = 1;
    }
    else {
     current->next = new ListNode(l1->val + n);
     current = current->next;
     n = 0;
    }
    l1 = l1->next;
   }
   else {
    if (l1->val + l2->val+n >= 10) {
     current->next = new ListNode(l1->val + l2->val + n-10);
     current = current->next;
     n = 1;
    }
    else {
     current->next = new ListNode(l1->val + l2->val + n);
     current = current->next;
     n = 0;
    }
    l1 = l1->next;
    l2 = l2->next;
   }

}
return a;
}

};

这题的内容就是利用列表来存储两个整数让后相加，将结果放到另外一个列表中，我采用的方法是用一个int变量n来储存是否需要进位，让后对提供的两个列表逐项相加（包括与n），然后同时指向下一项，储存的时候将每次的结果都用new一个结构体来储存。其中每次相加考虑的情况是两个都没到最后一项，两个中某一个到最后一项，两个都到最后一项但有进位。