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