2. Add Two Numbers

2. Add Two Numbers

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

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 题意一开始还真没看懂，模拟了下，才知道是加法。

直接模拟竖位运算加法，刚好这里也是反着的，加法每次就从最低位开始。如果链表长度不齐，到null了，就赋值0,。最后一位如果进位，别忘记特殊处理，再加一个链表。

迭代和递归都可以做。

迭代：

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        int carry = 0;
        ListNode tail = new ListNode(0);
        ListNode ptr = tail;
        while(l1!=null || l2!=null) {
            int tmp1 = 0;//必须要有，避免长度不一样
            if(l1!=null) {
                tmp1 = l1.val;
                l1 = l1.next;
            }
            int tmp2 = 0;//同上
            if(l2!=null) {
                tmp2 = l2.val;
                l2 = l2.next;
            }
            int tmp = tmp1 + tmp2 + carry;
            carry = tmp/10;
            ptr.next = new ListNode(tmp%10);
            ptr = ptr.next;
        }
        //如果两个链表都为空了，但最后一位进了1的情况
        if(carry==1) {
            ptr.next = new ListNode(1);
        }
        return tail.next;
    }
}

 递归做法：

public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        return helper(l1,l2,0);
    }

    public ListNode helper(ListNode l1, ListNode l2, int carry){
        if(l1==null && l2==null){
            return carry == 0? null : new ListNode(carry);
        }
        if(l1==null && l2!=null){
            l1 = new ListNode(0);
        }
        if(l2==null && l1!=null){
            l2 = new ListNode(0);
        }
        int sum = l1.val + l2.val + carry;
        ListNode curr = new ListNode(sum % 10);
        curr.next = helper(l1.next, l2.next, sum / 10);
        return curr;
    }
}