LeetCode2 Add Two Numbers

标签：leetcode   add two numbers   

题目：

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

分析：此题可以用归并算法，但是有两点需要注意，1：如何记录首结点。。2：如何保持链表中的串接关系。。见代码注释。

代码1：

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
		if(l1 == NULL) return l2;
		if(l2 == NULL) return l1;
		ListNode *pNode = NULL, *pNext = NULL, *pFirst = NULL;
        while(l1 != NULL && l2 != NULL){
            int value = (l1->val + l2->val);
			if(pNext == NULL)    // 不能进位
				pNext = new ListNode(value%10);
            else{  // 有向前进位
				pNext->val = (1+value)%10;
                value = value + 1;
            }
            if(value >= 10)  // 大于10 向前进一位
				pNext->next = new ListNode(1);
			if(pFirst == NULL) pFirst = pNode = pNext;    // 存首结点
			else{
				pNode->next = pNext;   // 保持串接关系
				pNode = pNext;
			}
			pNext = pNext->next;
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1 != NULL){
            int value = l1->val;
			if(pNext == NULL)
				pNext = new ListNode(value%10);
            else{  // 有向前进位
				pNext->val = (1+value)%10;
                value = value + 1;
            }
            if(value >= 10)  // 大于10 向前进一位
				pNext->next = new ListNode(1);
			
			pNode->next = pNext;   // 保持串接关系
			pNode = pNext;
			pNext = pNext->next;
            l1 = l1->next;
        }
        while(l2 != NULL){
            int value = l2->val;
			if(pNext == NULL)
				pNext = new ListNode(value%10);
            else{  // 有向前进位
				pNext->val = (1+value)%10;
                value = value + 1;
            }
            if(value >= 10)  // 大于10 向前进一位
				pNext->next = new ListNode(1);

			pNode->next = pNext;   // 保持串接关系
			pNode = pNext;

			pNext = pNext->next;
            l2 = l2->next;
        }
		return pFirst;
    }

ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
		ListNode *result = new ListNode(0);
		ListNode *pNode = result;
		bool flag = true;
        while(l1 != NULL && l2 != NULL){
            int value = (l1->val + l2->val);
			if(result->next == NULL)
				result->next = new ListNode(value%10);
            else{  // 有向前进位
                result->next->val = (1+value)%10;
                value = value + 1;
            }
            if(value >= 10)  // 大于10 向前进一位
				result->next->next = new ListNode(1);
            result = result->next; 
            l1 = l1->next;
            l2 = l2->next;
        }
        while(l1 != NULL){
            int value = l1->val;
			if(result->next == NULL)   // 没有向前进位
				result->next = new ListNode(value%10);
            else {    
                result->next->val = (1+value)%10;
                value = value + 1;
            }
            if(value >= 10)
				result->next->next = new ListNode(1);

            result = result->next;
            l1 = l1->next;
        }
        while(l2 != NULL){
            int value = l2->val;
			if(result->next == NULL)
				result->next = new ListNode(value%10);
            else {
                result->next->val = (1+value)%10;
                value = value + 1;
            }
            if(value >= 10)
				result->next->next = new ListNode(1);

            result = result->next;
            l2 = l2->next;
        }
		return pNode->next;
    }

LeetCode2 Add Two Numbers

标签：leetcode   add two numbers   

原文地址：http://blog.csdn.net/lu597203933/article/details/44875613