Add Two Numbers

全栈程序员站长

发布于 2022-07-05 10:38:08

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发布于 2022-07-05 10:38:08

You are given two linked lists representing two non-negative numbers. 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.

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

错误解法：由于数太长。用int，long存不下，导致结果错误。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null) return l2;
        if(l2==null) return l1;
        ListNode node1=l1;
        ListNode node2=l2;
        int a =0,b=0;
        int c=1;
        while(node1!=null) {
        	a+=(c*node1.val);
        	c=c*10;
        	node1=node1.next;
        }
        c=1;
        while(node2!=null){
        	b+=(c*node2.val);
        	c=c*10;
        	node2=node2.next;
        }
        int d=a+b;
        
        String res=Integer.toString(d);
        ListNode l3=new ListNode(0);
        ListNode node3=l3;
        for(int i=res.length()-1;i>=0;i--){
        	node3.val=(int)(res.charAt(i))-(int)('0');
        	if(i>0){
        		node3.next=new ListNode(0);
        		node3=node3.next;
        	}
        }
        return l3;
    }
}

正确解法。空间效率极差。肯定还有更好的解法，没工夫研究，待续……

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
     public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if(l1==null) return l2;
        if(l2==null) return l1;
        ListNode node1=l1;
        ListNode node2=l2;
        List<Integer> list1=new ArrayList<Integer>();
        List<Integer> list2=new ArrayList<Integer>();
        
        while(node1!=null) {
            list1.add(node1.val);
            node1=node1.next;
        }
       
        while(node2!=null){
            list2.add(node2.val);
            node2=node2.next;
        }
        
        while(list1.size()<list2.size()) list1.add(0);
        while(list2.size()<list1.size()) list2.add(0);
        
        ListNode l3=new ListNode(0);
        ListNode node3=l3;
        int tmp=0;
        for(int i=0;i<list1.size();i++){
            int a=list1.get(i)+list2.get(i)+tmp;
            if(a>=10){
                node3.val=a-10;
                tmp=1;
            }
            else{
                node3.val=a;
                tmp=0;
            }
            if(i<list1.size()-1){
                node3.next=new ListNode(0);
                node3=node3.next;
            }
        }
        
        if(tmp==1){
            node3.next=new ListNode(tmp);
        }
        return l3;
    }
}

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原始发表：2021年9月7，如有侵权请联系 cloudcommunity@tencent.com 删除

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