221. Add Two Numbers II【medium】

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in forward order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

 

Example

Given 6->1->7 + 2->9->5. That is, 617 + 295.

Return 9->1->2. That is, 912.

解法一：

class Solution {
public:
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    ListNode *addLists2(ListNode *l1, ListNode *l2) {
        reverse(l1);
        reverse(l2);
        ListNode *dummy = new ListNode(0);
        ListNode *tail = dummy;
        int carry = 0;

        while (l1 != NULL || l2 != NULL) {
            int sum = carry;
            if (l1 != NULL) {
                sum += l1->val;
                l1 = l1->next;
            }
            if (l2 != NULL) {
                sum += l2->val;
                l2 = l2->next;
            }
            if (sum > 9) {
                carry = 1;
                sum -= 10; 
            } else {
                carry = 0;
            }
            tail->next = new ListNode(sum);
            tail = tail->next;
        }

        if (carry == 1) {
            tail->next = new ListNode(1);
            tail = tail->next;
        }

        reverse(dummy->next);

        return dummy->next;
    }


    void reverse(ListNode *&head) {
        ListNode *prev = NULL;

        while (head != NULL) {
            ListNode *temp = head->next;
            head->next = prev;
            prev = head;
            head = temp;
        }

        head = prev;
    }
};

链表先翻转，再求和，然后再翻转

解法二：

public class Solution {  
    /** 
     * @param l1: the first list 
     * @param l2: the second list 
     * @return: the sum list of l1 and l2  
     */  
    public ListNode addLists2(ListNode l1, ListNode l2) {  
        Stack<Integer> temp1 = reverseNode(l1);  
        Stack<Integer> temp2 = reverseNode(l2);  

        ListNode point = new ListNode(0);  
        int flag = 0;  

        while((!temp1.isEmpty()) && (!temp2.isEmpty())){  
            int value = temp1.pop() + temp2.pop() + flag;  
            flag = value / 10;  
            value = value % 10;  
            ListNode cur = new ListNode(value);  
            cur.next = point.next;  
            point.next = cur;  
        }  

        while(!temp1.isEmpty()){  
            int value = temp1.pop() + flag;  
            flag = value / 10;  
            value = value % 10;  
            ListNode cur = new ListNode(value);  
            cur.next = point.next;  
            point.next = cur;  
        }  

        while(!temp2.isEmpty()){  
            int value = temp2.pop() + flag;  
            flag = value / 10;  
            value = value % 10;  
            ListNode cur = new ListNode(value);  
            cur.next = point.next;  
            point.next = cur;  
        }  

        if(flag == 1){  
            ListNode cur = new ListNode(1);  
            cur.next = point.next;  
            point.next = cur;  
        }  

        return point.next;  
    }  
     
 
    public Stack<Integer> reverseNode(ListNode temp){  
        Stack<Integer> record = new Stack<Integer>();  

        while(temp != null){  
            record.push(temp.val);  
            temp = temp.next;  
        }  

        return record;  
    }  
}  

利用栈的先进后出，记录两个链表的节点值。然后计算对应位置的两个数字之和，如果有进位，赋值给flag并带入下一个计算。

参考@sunday0904 的代码