LeetCode#445. Add Two Numbers II

• 题目：给定两个非空链表，表示两个非负整数，链表的第一个元素存储整数的最高位，链表的每个节点表示整数的一位。将两个链表表示的整数相加，并且以链表的形式返回相加结果。
• 难度：Medium
• 思路：整数相加是先从低位开始，而我们的链表表头存储的是高位，因为无法通过直接遍历链表得到相加结果。可借助栈来实现先访问链表的低位。想到栈可以实现，便也可通过递归来实现。

• 代码：

  借助栈实现
  在相加的过程中，构建链表。

    class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> s1 = new Stack<Integer>();
        Stack<Integer> s2 = new Stack<Integer>();

        while(l1 != null) {
            s1.push(l1.val);
            l1 = l1.next;
        };
        while(l2 != null) {
            s2.push(l2.val);
            l2 = l2.next;
        }

        int sum = 0;
        ListNode list = new ListNode(0);
        while (!s1.empty() || !s2.empty()) {
            if (!s1.empty()) sum += s1.pop();
            if (!s2.empty()) sum += s2.pop();
            list.val = sum % 10;
            ListNode head = new ListNode(sum / 10);
            head.next = list;
            list = head;
            sum /= 10;
        }

        return list.val == 0 ? list.next : list;
    }
}

递归实现

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int len1 = getLength(l1);
        int len2 = getLength(l2);
        ListNode head = new ListNode(0);

        if (len1 < len2){
            head.next = helper(l2, l1, len2-len1);
        }else{
            head.next = helper(l1, l2, len1-len2);
        }
        if(head.next.val > 9){
            head.next.val = head.next.val%10;
            head.val += 1;
            return head;
        }

        return head.next;

    }

    public int getLength(ListNode node){
        int count = 0;
        while(node != null){
            count++;
            node = node.next;
        }
        return count;
    }

    public ListNode helper(ListNode l1, ListNode l2, int offset){
        if (l1 == null){
            return null;
        }
        ListNode result = null;
        ListNode post = null;
        if (offset == 0){
            result = new ListNode(l1.val + l2.val);
            post = helper(l1.next, l2.next, 0);
        }else{
            result = new ListNode(l1.val);
            post = helper(l1.next, l2, offset-1);
        }

        if(post != null && post.val > 9){
            result.val += 1;
            post.val = post.val%10;
        }
        result.next = post;
        return result;


    }
}

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