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 Subscribe to see which companies asked this question
题意:实际上就是342+465=807然后倒序输出。【一开始也没有看明白。】
参考别人的。一开始题意都没看明白。。。。。。
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) { struct ListNode *l3 = (struct ListNode *)malloc(sizeof(struct ListNode)); l3->val=0; l3->next=NULL; //这边因为l3是一直往后遍历的,所以返回头的话要加个head指示一下。 struct ListNode* head = l3; l3->val=l1->val+l2->val; //while循环判断是否申请新的节点 l1=l1->next; l2=l2->next; while(l1||l2||l3->val>9){ l3->next=(struct ListNode *)malloc(sizeof(struct ListNode)); l3->next->val=0; l3->next->next=NULL; if(l1){ l3->next->val+=l1->val; l1=l1->next; } if(l2){ l3->next->val+=l2->val; l2=l2->next; } if(l3->val>9){ l3->next->val+=l3->val/10; l3->val=l3->val%10; } l3=l3->next; } return head; }
PS:主要是搞清楚有几种情况。9+8这种需要申请节点;12+9这种不等长的;
注意C语言链表的使用。。。
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode addTwoNumbers(ListNode l1, ListNode l2) { ////直接模拟加法就行。先申请一个head ListNode head=new ListNode(0); int carry=0; ListNode p1=l1,p2=l2,p3=head; while(l1!=null || l2!=null){ if(l1!=null){ carry+=l1.val; l1=l1.next; } if(l2!=null){ carry+=l2.val; l2=l2.next; } p3.next=new ListNode(carry%10); p3=p3.next; carry=carry/10; } if(carry==1){ p3.next=new ListNode(1); } return head.next; } }
JAVA版本的。