LeetCode OJ (2)

来源：互联网发布：剑灵龙男捏脸数据大全编辑：程序博客网时间：2024/06/14 12:49

Add Two Numbers

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

初始代码块（初学者水平）

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {    ListNode *header=NULL,*sum=NULL;    int carry=0;    bool flag=false;    while(l1!=NULL && l2!=NULL){        ListNode *tmp=new ListNode(0);        carry=l1->val+l2->val+flag;        if(carry>=10){            tmp->val = carry-10;            flag = true;        }        else{            tmp->val=carry;            flag=false;        }        if(!header)            header=tmp;        else            sum->next=tmp;        sum=tmp;        l1=l1->next;        l2=l2->next;    }    ListNode *last;    if(l1==NULL)        last=l2;    else        last=l1;    carry=0;    while(last!=NULL){        ListNode *tmp=new ListNode(0);        carry=last->val+flag;        if(carry>=10){           tmp->val=carry-10;            flag=true;        }        else{            tmp->val=carry;            flag=false;        }        sum->next=tmp;        sum=tmp;        last=last->next;    }    if(flag){        ListNode *tmp=new ListNode(0);        tmp->val=flag;        sum->next=tmp;    }    return header;}}

经过优化和学弟提醒之后

ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {    ListNode *header=NULL,*sum=NULL;    int carry=0;    while(l1 || l2){        new ListNode(0);        if(l1)  carry+=l1->val;        if(l2)  carry+=l2->val;        ListNode *tmp=new ListNode(carry%10);        if(!header)            header=tmp;        else            sum->next=tmp;        carry/=10;        sum=tmp;        if(l1) l1=l1->next;        if(l2) l2=l2->next;    }    if(carry){    ListNode *tmp=new ListNode(carry);    sum->next=tmp;    }    return header;}

我太悲伤了，第二个代码块是我奋斗的目标！

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