leetcode Add two numbers-优快云博客

本文深入探讨了在Visual Studio 2010环境中声明指针时未赋值与赋值为NULL的区别，并通过实例展示了这种差异在LeetCode平台上的具体表现与解决策略。

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在vs2010里，指针声明的时候不赋值也不置NULL是没问题的。不过leetcode上如果不赋值，必须初始化为NULL，不然就RE。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
        ListNode *res = NULL;
    	if(!l1){
    		res = l2;return res;
    	}
    	if(!l2){
    		res = l1;return res;
    	}
    
    	ListNode *p = NULL;
    	int carry = 0, tmp;
    	while(l1 && l2){
    		tmp = l1->val + l2->val + carry;
    		carry = tmp/10;
    		tmp %= 10;
    		if(!p){
    			p = new ListNode(tmp);
    			res = p;
    		}else{
    			ListNode *ntmp = new ListNode(tmp);
    			p->next = ntmp;
    			p = ntmp;
    		}
    		l1 = l1->next;
    		l2 = l2->next;
    	}
    	while(l1){
    		tmp = l1->val + carry;
    		carry = tmp/10;
    		tmp %= 10;
    		if(!p){
    			p = new ListNode(tmp);
    			res = p;
    		}else{
    			ListNode *ntmp = new ListNode(tmp);
    			p->next = ntmp;
    			p = ntmp;
    		}
    		l1 = l1->next;
    	}
    	while(l2){
    		tmp = l2->val + carry;
    		carry = tmp/10;
    		tmp %= 10;
    		if(!p){
    			p = new ListNode(tmp);
    			res = p;
    		}else{
    			ListNode *ntmp = new ListNode(tmp);
    			p->next = ntmp;
    			p = ntmp;
    		}
    		l2 = l2->next;
    	}
    	if (carry>0)
    	{
    		ListNode *ntmp = new ListNode(carry);
    		p->next = ntmp;
    	}
    
    	return res;
    }
};

不过还是有一些更简洁的写法：

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode current = new ListNode(0);
        ListNode head = current;

        int shift = 0;
        do
        {
            current.val = ((l1!=null)?l1.val:0) + ((l2!=null)?l2.val:0) + shift;
            shift = current.val / 10;
            current.val%=10;
            l1 = (l1!=null)?l1.next:null;
            l2 = (l2!=null)?l2.next:null;
            if((l1==null)&&(l2==null)){break;}

            current.next = new ListNode(0);
            current = current.next;
        }while(l1!=null || l2!=null);
        if(shift>0){current.next = new ListNode(1);}
        return head;
    }