AddTwoSum

Example:

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

 1 class ListNode {
 2     int val;
 3     ListNode next;
 4 
 5     ListNode(int x) {      //对应结构体初始化为ListNode(int data=0,struct ListNode next=NULL):val(val),next(next){}
 6         val = x;
 7         next = null;
 8     }
 9 }
10 
11 class Solution5 {
12     ListNode addTwoNumbers(ListNode l1, ListNode l2) {
13         ListNode preHead = new ListNode(0);
14         ListNode p = l1, q = l2, curr = preHead;
15         int carry = 0;
16         while (p != null || q != null) {
17             int x = (p != null) ? p.val : 0;
18             int y = (q != null) ? q.val : 0;
19             int sum = x + y + carry;                //进位处理，并在下一个节点处进行处理
20             carry = sum / 10;
21             curr.next = new ListNode(sum % 10); //新构造一个节点
22             curr = curr.next;
23             if (p != null) p = p.next;
24             if (q != null) q = q.next;
25         }
26         if (carry > 0) {                            //处理多一位
27             curr.next = new ListNode(carry);
28         }
29         return preHead.next;
30     }
31 }

此题基本就是用链表处理大数，所以用int+ListNode的方式来处理大数问题，只是应该找到其适用场景，譬如较多的写操作、内存占用等。

 1 #include <bits/stdc++.h>
 2 #include <map>
 3 #include <vector>
 4 using namespace std;
 5 
 6 struct ListNode
 7 {
 8     int val;
 9     struct ListNode *next;
10     ListNode(int data = 0, struct ListNode *next = NULL) : val(val), next(next) {} //初始化链表
11 };
12 
13 class Solution
14 {
15   public:
16     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2)
17     {
18         ListNode preHead(0), *p = &preHead;
19         int carry = 0;
20         while (l1 || l2)
21         {
22             int sum = (l1 ? l1->val : 0) + (l2 ? l2->val : 0) + carry;
23             carry = sum / 10;
24             p->next = new ListNode(sum % 10);
25             p = p->next;
26             l1 = l1 ? l1->next : l1;
27             l2 = l2 ? l2->next : l2;
28         }
29         if (carry > 0)
30         { //处理多一位
31             p->next = new ListNode(carry);
32         }
33         return preHead.next;
34     }

 

posted @ 2018-10-08 10:09  sigmod  阅读(133)  评论(0)  编辑  收藏  举报