leetcode2-两数相加

原题

给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。

如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外,这两个数都不会以 0 开头。

示例:

输入: (2 -> 4 -> 3) + (5 -> 6 -> 4)
输出: 7 -> 0 -> 8
原因: 342 + 465 = 807

解法

思想

模拟数学规则,将每一位对应的值和进位相加,将null一般化为0。

代码

Java

/**
 * 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) {
        ListNode preHead = new ListNode(0);
        int val1,val2,sum = 0;
        int carry = 0;
        ListNode cur = preHead;
        while(l1!=null || l2!=null || carry != 0){
            val1 = l1==null?0:l1.val;
            val2 = l2==null?0:l2.val;
            sum = val1+val2+carry;
            if(sum>=10) {
                carry = 1;
                sum -= 10;    
            }else carry = 0;
            cur.next = new ListNode(sum);
            cur = cur.next;
            l1 = l1 == null?null:l1.next;
            l2 = l2 == null?null:l2.next;
        }
        return preHead.next;
    }
}

Go(2020/07/04)

func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
    var dummy *ListNode = new(ListNode)
    var node *ListNode = dummy
    var num1,num2,carry int
    for{
        if l1 == nil && l2 == nil && carry == 0 { return dummy.Next }
        if l1 != nil{
            num1, l1 = l1.Val, l1.Next
        }else{
            num1 = 0
        } 
        if l2 != nil{
            num2, l2 = l2.Val, l2.Next
        }else{
            num2 = 0
        }
        sum := num1 + num2 + carry
        if sum >= 10{
            sum,carry =  sum - 10 , 1
        }else{
            carry =  0
        }
        node.Next = &ListNode{Val:sum}
        node= node.Next
    }
}

Python(2020/07/04)

(秉承人生苦短原则,Python条件分支尽量都写在了一行,如果不清晰的话请看上面语言的写法)

class Solution(object):
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        cur = dummy
        num1 = num2 = carry = 0
        while l1 != None or l2 != None or carry != 0:
            if l1 != None: num1, l1 = l1.val, l1.next
            else: num1 = 0
            if l2 != None: num2, l2 = l2.val, l2.next
            else: num2 = 0
            sum = num1 + num2 + carry
            if sum >= 10:sum,carry =  sum - 10 , 1
            else: carry =  0
            cur.next = ListNode(sum)
            cur = cur.next
        return dummy.next

原创文章,作者:彭晨涛,如若转载,请注明出处:https://www.codetool.top/article/leetcode2-%e4%b8%a4%e6%95%b0%e7%9b%b8%e5%8a%a0/