2. Add Two Numbers

最新推荐文章于 2021-12-02 14:05:46 发布

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最新推荐文章于 2021-12-02 14:05:46 发布

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本文链接：https://blog.youkuaiyun.com/stonesola/article/details/79820013

本文介绍如何使用链表来解决两个非负整数相加的问题，通过将整数以逆序方式存储在链表中，实现了进位处理及最终结果的链表返回。提供Java和Python两种语言的实现方案。

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问题

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.

Example

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

分析：关键是解决进位问题，val = l1.val + l2.val + 进位。需要判断l1和l2是否是结尾，对结尾情况作出处理。最后一位进位需要在末尾加一个链。

PS：java需要引用实体对象才能算是真正的引用

java

开始的做法：

/**
 * 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) {
        int temp = 0;
        ListNode res = new ListNode(0);
        ListNode now = res, p = l1, q = l2;
        while(p != null && q != null){
            int val = p.val + q.val + temp;
            now.next = new ListNode(val%10);
            temp = val / 10;
            now = now.next;
            p = p.next;
            q = q.next;
        }
        ListNode lastL = null;
        if(p != null){
            lastL = p;
        }
        else if(q != null){
            lastL = q;
        }
        while(lastL != null){
            int val = lastL.val + temp;
            now.next = new ListNode(val%10);
            temp = val / 10;
            now = now.next;
            lastL = lastL.next;
        }

        if(temp != 0){
            now.next = new ListNode(temp);
        }
        return res.next;
    }
}

简化代码，可以在两个链表相加时，空链表当前值设置为0：

/**
 * 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) {
        int temp = 0, val, pval, qval;
        ListNode res = new ListNode(0);
        ListNode now = res, p = l1, q = l2;
        while(p != null || q != null){
            pval = p != null ? p.val : 0;
            qval = q != null ? q.val : 0;
            val = pval + qval + temp;
            now.next = new ListNode(val % 10);
            temp = val / 10;
            now = now.next;
            if(p != null) p = p.next;
            if(q != null) q = q.next;
        }

        if(temp != 0){
            now.next = new ListNode(temp);
        }
        return res.next;
    }
}

Python

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

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        res = ListNode(0)
        now = res
        p = l1
        q = l2
        carry = 0 #进位
        while p or q:
            pval = p.val if p else 0
            qval = q.val if q else 0
            val = carry + pval + qval
            now.next = ListNode(val % 10)
            carry = val // 10
            now = now.next
            if p: p = p.next
            if q: q = q.next
        if carry:
            now.next = ListNode(carry)

        return res.next