Leetcode 2. 两数相加

2. 两数相加

给你两个非空的链表，表示两个非负的整数。它们每位数字都是按照 逆序 的方式存储的，并且每个节点只能存储 一位 数字。

请你将两个数相加，并以相同形式返回一个表示和的链表。

你可以假设除了数字 0 之外，这两个数都不会以 0 开头。

方法一：假设两个链表至少有一个不空

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        ListNode node = new ListNode(0);
        ListNode one = l1, two = l2, start = node;
        int sum = 0;

        while (one!= null || two!= null){
            sum /= 10;
            if(one != null){
                sum += one.val;
                one = one.next;
            }

            if(two != null){
                sum += two.val;
                two = two.next;
            }

            start.next = new ListNode(sum%10);
            start = start.next;
        }

        if(sum/10 != 0) start.next = new ListNode(sum/10);
        return node.next;
    }

方法二：假设两个链表必须都不空

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

        ListNode node = new ListNode(0);
        ListNode one = l1, two = l2, start = node;
        int sum = 0;

        while (one!= null && two!= null){
            sum /= 10;

            sum = sum+ one.val+two.val;
            one = one.next;
            two = two.next;


            start.next = new ListNode(sum%10);
            start = start.next;
        }

        if(one != null){
            while (one != null){
                sum /= 10;
                sum = sum+one.val;
                one = one.next;
                start.next = new ListNode(sum%10);
                start = start.next;

            }

        }
        else {
            while (two != null){
                sum /= 10;
                sum = sum+two.val;
                two = two.next;
                start.next = new ListNode(sum%10);
                start = start.next;
            }
        }

        if(sum/10 != 0) start.next = new ListNode(sum/10);
        return node.next;

    }

方法三：递归

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return dfs(l1, l2, 0);
    }

    ListNode dfs(ListNode l, ListNode r, int i) {
        if (l == null && r == null && i == 0) return null;
        int sum = (l != null ? l.val : 0) + (r != null ? r.val : 0) + i;
        var node = new ListNode(sum % 10);
        if(l != null) l = l.next;
        if(r != null) r = r.next;

        node.next = dfs(l, r, sum / 10);
        return node;
    }