2. Add Two Numbers

2. Add Two Numbers

Medium

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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.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

错误解法

/**
 * 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) {
      String s1 = "";
        while(l1 != null){
            int val = l1.val;
            s1 = val + s1;
            l1 = l1.next;
        }


        String s2 = "";
        while(l2 != null){
            int val = l2.val;
            s2 = val + s2;
            l2 = l2.next;
        }


        int result = Integer.parseInt(s1) + Integer.parseInt(s2);
  if(result == 0){
            return new ListNode(0);
        }
        ListNode head = new ListNode(1);
        ListNode cur = head;
        while(result != 0){
            int temp = result % 10 ;
            cur.next = new ListNode(temp);
           cur = cur.next;
            result /= 10;
        }
        return head.next;

    }
}

/**
 * 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) {
        if(l1 == null && l2 == null)
            return null;
        if(l1.val == 0 && l2.val == 0)
            return new ListNode(0);
        int n1 = 0;
        int n2 = 0;
        ArrayList<Integer> list1 = new ArrayList<>();
        ArrayList<Integer> list2 = new ArrayList<>();
        
        while(l1 != null){
            list1.add(l1.val);
            l1 = l1.next;
        }
        while(l2 != null){
            list2.add(l2.val);
            l2 = l2.next;
        }
    
        for(int i = list1.size() - 1; i >= 0; i--){
            n1 = 10 * n1 + list1.get(i);
        }
        for(int i = list2.size() - 1; i >= 0; i--){
            n2 = 10 * n2 + list2.get(i);
        }
        
        int ret = n1 + n2;
        if(ret == 0)
            return null;
        ListNode head =  new ListNode(1);
        ListNode newHead = head;
        
        while(ret > 0){
            int d = ret % 10;
            ret /= 10;
            head.next = new ListNode(d);
            head = head.next;
        }
        return newHead.next;
    }
}

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正确解法

不用转换，直接读取一个，计算一个

/**
 * 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 head = new ListNode(0);
        ListNode cur = head;
        int  carry = 0;
        while(l1 != null || l2 != null){
           int x = (l1 != null) ? l1.val : 0;
           int y = (l2 != null) ? l2.val : 0;
           int sum = x + y + carry;
            cur.next = new ListNode(sum % 10) ;
            cur = cur.next;
            carry = sum / 10;//进位永远为1
            if(l1 != null)//l1和l2可能长度不一致
                l1 = l1.next;
            if(l2 != null)
                l2 = l2.next;
        }
        if(carry > 0)
            cur.next = new ListNode(carry);
        return head.next;
        
    }
}