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  • Leetcode 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

    考查的是链表的合并,类似多项式的合并,开始打算利用链表本身的空间去做,但考虑到两个链表共享同一个空间,会导致断链,故还是采取了增加空间复杂度

    如果数据量小的话,可以先把链表转换成整数,然后相交后再转换成链表的形式

    #include <iostream>
    #include <algorithm>
    
    using namespace std;
    
    static int debug = 0;
    
    struct ListNode{
        int val;
        ListNode *next;
        ListNode(int x):val(x),next(NULL){}
    };
    
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2){
        if (l1 == NULL) return l2;
        if (l2 == NULL) return l1;
        ListNode *head = NULL;
        ListNode *l3 = head;
        
        int carry = 0;
        while (l1 != NULL && l2 != NULL) {
            int sum = l1->val + l2->val + carry;
            carry = sum/10;
            sum = sum%10;
            if (head == NULL) {
                head = new ListNode(sum);
                l3 = head;
            }else{
                ListNode *tmp = new ListNode(sum);
                l3 ->next = tmp;
                l3 = tmp;
            }
            l1 = l1->next;
            l2 = l2->next;
        }
        while (l1!=NULL) {
            int sum = carry + l1->val;
            carry = sum/10;
            sum = sum%10;
            ListNode *tmp = new ListNode(sum);
            l3 ->next = tmp;
            l3 = tmp;
            l1 = l1->next;
        }
        while (l2!=NULL) {
            int sum = carry + l2->val;
            carry = sum/10;
            sum = sum%10;
            ListNode *tmp = new ListNode(sum);
            l3 ->next = tmp;
            l3 = tmp;
            l2 = l2->next;
        }
        if (carry) {
            ListNode *tmp = new ListNode(carry);
            l3 ->next = tmp;
            l3 = tmp;
        }
        return head;
    }
    
    
    int main(){
        ListNode *a = new ListNode(5);
        ListNode *b = new ListNode(5);
        
        ListNode *c = addTwoNumbers(a,b);
        while (c != NULL) {
            cout<<c->val<<endl;
            c = c->next;
        }
    }
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  • 原文地址:https://www.cnblogs.com/xiongqiangcs/p/3626132.html
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