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.
輸入為兩個(gè)非空鏈表代表兩個(gè)非負(fù)的整數(shù)。這些整數(shù)以逆序的形式存儲(chǔ)，并且每個(gè)元素包含一個(gè)一位數(shù)。將這兩個(gè)整數(shù)相加得到一個(gè)新的鏈表并返回。
可以假定兩個(gè)整數(shù)沒有前導(dǎo)0，除了整數(shù)0本身。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

samples

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

用鏈表運(yùn)算來表示加法，即342+465=807

Solutions

<ol>
<li> O(n)

// com.cnsumi.leetcode.algorithms. A0002_Add_Two_Numbers.java;
class ListNode {
    int val;
    ListNode next;
    public ListNode(int val) {
        this.val = val;
    }
    
    @Override
    public String toString() {
        StringBuilder ret = new StringBuilder();
        ret.append("(");
        ListNode node = this;
        do {
            if (node != this) {
                ret.append("->");   
            }
            ret.append(node.val);
            node = node.next;
        } while (node != null);
        ret.append(")");
        return ret.toString();
    }
}

public class A0002_Add_Two_Numbers {
    public static void main(String[] args) {
        ListNode l1 = new ListNode(2);
        l1.next = new ListNode(4);
        l1.next.next = new ListNode(3);
        ListNode l2 = new ListNode(5);
        l2.next = new ListNode(6);
        l2.next.next = new ListNode(4);
        System.out.println(l1);
        System.out.println(l2);
        System.out.println(solution1(l1, l2));
    }
    
    public static ListNode solution1(ListNode l1, ListNode l2) {
        int carry = 0;
        ListNode head = new ListNode(0);
        ListNode pre = head;
        while (l1 != null || l2 != null || carry > 0) {
            int sum = (l1 == null ? 0 : l1.val) + (l2 == null ? 0 : l2.val) + carry;
            carry = sum / 10;
            pre.next = new ListNode(sum % 10);
            pre = pre.next;
            l1 = l1 == null ? l1 : l1.next;
            l2 = l2 == null ? l2 : l2.next;
        }
        return head.next;
    }
}