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

 

两个数存在两个单链表中，求他们的和，结果还是存在一个链表中。

 

Java程序

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode headNode = new ListNode(0);  
        ListNode currentNode = headNode;  
        
        int carray = 0;  
        while(l1 != null && l2 != null){  
                carray +=l1.val;
                l1 = l1.next;
                
                carray +=l2.val;
                l2 = l2.next;

           currentNode.next = new ListNode(carray%10);
           currentNode = currentNode.next;
           carray/=10;
           
        }
        if(l1!=null){
           
                while(l1!=null){
                    carray +=l1.val;
                    l1 = l1.next;
                    currentNode.next = new ListNode(carray%10);
                    currentNode = currentNode.next;
                    carray/=10;
                }
            }
            
        if(l2!=null){
                while(l2!=null){
                    carray +=l2.val;
                    l2 = l2.next;
                    currentNode.next = new ListNode(carray%10);
                    currentNode = currentNode.next;
                    carray/=10;
                }
            }
        if(carray==1){
            currentNode.next = new ListNode(1);
            currentNode = currentNode.next;
        }
       return headNode.next;
    }
}

第一两个节点，一个指向头节点，一个指向当前运行节点

同时遍历两个单链表，对应位置求和，carray%10为新的节点值，carray/10是来进位的

当有一个链表是空的时候说明是两个长度不同的数相加，对当前节点再考虑进位的情况下，将非空的节点链接起来。

 

上面的while是当两个链表都不为空的时候在进行运算，下面再连接上非空的链表

 

可把这两个组合在一起

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode headNode = new ListNode(0);  
        ListNode currentNode = headNode;  
        
        int carray = 0;  
        while(l1 != null || l2 != null){  
            if(l1!=null){
                carray +=l1.val;
                l1 = l1.next;
            }
            if(l2!=null){
                carray +=l2.val;
                l2 = l2.next;
            }
           currentNode.next = new ListNode(carray%10);
           currentNode = currentNode.next;
           carray/=10;
           
        }
        
        if(carray==1){
            currentNode.next = new ListNode(1);
            currentNode = currentNode.next;
        }
       return headNode.next;
    }
}

这样程序简洁多了

 

Python程序

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        carry = 0 
        head = ListNode(0)
        currentNode = head
        while l1 or l2:
            if l1:
                carry +=l1.val
                l1 = l1.next
            if l2:
                carry +=l2.val
                l2 = l2.next
                
            currentNode.next = ListNode(carry%10)
            currentNode = currentNode.next
            carry = carry//10
            
        if carry==1:
            currentNode.next = ListNode(1)
            currentNode= currentNode.next
        return head.next