LeetCode002 Add Two Numbers C语言

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 -> 8Subscribe 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;        }        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);            p3=p3.next;            carry=carry/10;        }        if(carry==1){            p3.next=new ListNode(1);        }        return head.next;     }}

JAVA版本的。