Add Two Numbers I、II

Add Two Numbers I

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

思路

1. 从同时从2个linkedlist头部开始遍历，算2个数的和。
2. 怎么算？sum = (l1.val + l2.val + carry) % 10, carry = (l1.val + l2.val) / 10
3. 不要忘了，l1和l2的长度可能不等，所以当某一个已经遍历完以后，还需要把剩下的那个遍历完。
4. 最后不要忘了最后一个可能出现的carry。比如999 + 99 => 9->8->0->(1) 但是，如果carry是0时就不需要加了。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        } else if (l2 == null) {
            return l1;
        }
        
        int carry = 0;
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
      //  Queue queue = new LinkedList();
        
        while (l1 != null && l2 != null) {
            int sum = l1.val + l2.val + carry;
            int cur = sum % 10;
            carry = sum / 10;
            head.next = new ListNode(cur);
            head = head.next;
            l1 = l1.next;
            l2 = l2.next;
        }
        
        while (l1 != null) {
            int sum = l1.val + carry;
            int cur = sum % 10;
            carry = sum / 10;
            head.next = new ListNode(cur);
            head = head.next;
            l1 = l1.next;
        }
        
        while (l2 != null) {
            int sum = l2.val + carry;
            int cur = sum % 10;
            carry = sum / 10;
            head.next = new ListNode(cur);
            head = head.next;
            l2 = l2.next;
        }
        
        if (carry != 0) {
            head.next = new ListNode(carry);
        }
        
        return dummy.next;
    }
}

Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Follow up:
What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

思路

1. 不能翻转listNode的情况下，就只能借助外部结构来实现翻转linkedList了，那么可以用FILO的STACK来分别存储L1和L2
2. 同时从q1/q2中取出节点，算2者相加的结果（逻辑与Add Two Numbers一致），算好的结果仍然放入到第三个STACK结构中
3. 从存结果的stack中依次取出结果的每个节点，并组织成LinkedList。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        //1. 不能翻转listNode的情况下，就只能借助外部结构来实现翻转linkedList了，那么久可以用FILO的STACK来存储L1和L2
        Stack q1 = new Stack();
        Stack q2 = new Stack();
        
        while (l1 != null) {
            q1.add(l1);
            l1 = l1.next;
        }
        
        while (l2 != null) {
            q2.add(l2);
            l2 = l2.next;
        }
        
        //2. 同时从q1/q2中取出节点，算2者相加的结果（逻辑与Add Two Numbers一致），算好的结果仍然放入到STACK结构中
        Stack result = new Stack();
        int carry = 0;
        while (!q1.isEmpty() && !q2.isEmpty()) {
            int sum = q1.pop().val + q2.pop().val + carry;
            int cur = sum % 10;
            carry = sum / 10;
            result.add(new ListNode(cur));
        }
        
        while (!q1.isEmpty()) {
            int sum = q1.pop().val + carry;
            int cur = sum % 10;
            carry = sum / 10;
            result.add(new ListNode(cur));
        }
        
        while (!q2.isEmpty()) {
            int sum = q2.pop().val + carry;
            int cur = sum % 10;
            carry = sum / 10;
            result.add(new ListNode(cur));
        }
        
        if (carry != 0) {
            result.add(new ListNode(carry));
        }
        
        //3. 从stack中依次取出结果的每个节点，并组织成LinkedList
        ListNode dummy = new ListNode(0);
        ListNode head = dummy;
        
        while (!result.isEmpty()) {
            head.next = result.pop();
            head = head.next;
        }
        return dummy.next;
    }
}