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[LeetCode] 002.Add Two Numbers (

[LeetCode] 002.Add Two Numbers (

作者: QyQiaoo | 来源:发表于2017-08-20 17:02 被阅读0次

    Problem description

    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 & Output

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

    Code

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

    Analysis

    • 注意链表的建立过程。

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