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  • LeetCode-Add Two Numbers

    Add Two Numbers Nov 1 '11 5998 / 20033

    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

    可能出现一条链表比另一条链表长的情况,需要注意进位

     1 /**
     2  * Definition for singly-linked list.
     3  * struct ListNode {
     4  *     int val;
     5  *     ListNode *next;
     6  *     ListNode(int x) : val(x), next(NULL) {}
     7  * };
     8  */
     9 class Solution {
    10 public:
    11     ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
    12         // Start typing your C/C++ solution below
    13         // DO NOT write int main() function
    14    
    15         ListNode* head=new ListNode(0);
    16         ListNode* ret=head;
    17         bool carry=false;
    18         while(l1!=NULL&&l2!=NULL){
    19         int sum=l1->val+l2->val;
    20         if(carry){sum++;carry=false;}
    21         if(sum>9)
    22         {
    23         carry=true;
    24         sum-=10;
    25         }
    26         ret->next=new ListNode(sum);
    27         ret=ret->next;
    28          l1=l1->next;
    29          l2=l2->next;
    30         }
    31         while(l1!=NULL){
    32         int sum=l1->val;
    33         if(carry){sum++;carry=false;}
    34         if(sum>9)
    35         {
    36         carry=true;
    37         sum-=10;
    38         }
    39         ret->next=new ListNode(sum);
    40         ret=ret->next;
    41          l1=l1->next;
    42         }
    43 while(l2!=NULL){
    44         int sum=l2->val;
    45         if(carry){sum++;carry=false;}
    46         if(sum>9)
    47         {
    48         carry=true;
    49         sum-=10;
    50         }
    51         ret->next=new ListNode(sum);
    52         ret=ret->next;
    53          l2=l2->next;
    54         }
    55 
    56         if(carry){
    57             ret->next=new ListNode(1);
    58         }
    59         return head->next;
    60     }
    61 };
    代码
    /**
     * Definition for singly-linked list.
     * public class ListNode {
     *     int val;
     *     ListNode next;
     *     ListNode(int x) {
     *         val = x;
     *         next = null;
     *     }
     * }
     */
    public class Solution {
        public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
            // Note: The Solution object is instantiated only once and is reused by each test case.
            if(l1==null&&l2==null)return null;
            if(l1==null)return l2;
            if(l2==null)return l1;
            boolean carry=false;
            ListNode ret=new ListNode(-1);
            ListNode current=ret;
            while(l1!=null&&l2!=null){
                int val=l1.val+l2.val;
                if(carry){
                    val++;
                    carry=false;
                }
                if(val>=10){
                    carry=true;
                    val-=10;
                }
                current.next=new ListNode(val);
                current=current.next;
                l1=l1.next;
                l2=l2.next;
            }
            while(l1!=null){
                int val=l1.val;
                if(carry){
                    val++;
                    carry=false;
                }
                if(val>=10){
                    carry=true;
                    val-=10;
                }
                current.next=new ListNode(val);
                current=current.next;
                l1=l1.next;
            }
            while(l2!=null){
                int val=l2.val;
                if(carry){
                    val++;
                    carry=false;
                }
                if(val>=10){
                    carry=true;
                    val-=10;
                }
                current.next=new ListNode(val);
                current=current.next;
                l2=l2.next;
            }
            if(carry){
                current.next=new ListNode(1);
            }
            return ret.next;
        }
    }
    Java
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  • 原文地址:https://www.cnblogs.com/superzrx/p/3177503.html
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