Add two numbers (linked list)

You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse
order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.

Given 3->1->5 and 5->9->2, return 8->0->8.

我的解法是先将两个链表中的数字加到 sum 里,再将 sum 转化成链表。但是如果输入的链表代表的数字如果很大,超过 int 的范围,造成 overflow,那么就会出问题。

因此,应该一位一位实现加法,直接生成新的链表。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;      
 *     }
 * }
 */
public class Solution {
    /**
     * @param l1: the first list
     * @param l2: the second list
     * @return: the sum list of l1 and l2 
     */
    public ListNode addLists(ListNode l1, ListNode l2) {
        // write your code here
        if (l1 == null && l2 == null) {
            return null;
        }
            
        int sum = 0;
        int multiplier = 1;
        
        if (l1 != null) {
            
            ListNode temp = l1;
            while (temp != null) {
                sum += multiplier * temp.val;
                multiplier*=10;
                temp = temp.next;
            }
            
        }
        
        if (l2 != null) {
            multiplier = 1;
            ListNode temp = l2;
            
            while (temp != null) {
                sum += multiplier * temp.val;
                multiplier *= 10;
                temp = temp.next;
            }
        }
        
        ListNode head = new ListNode(0);
        ListNode tempNode = head;
        do {
            int value = sum % 10;
            tempNode.next = new ListNode(value);
            sum = sum / 10;
            tempNode = tempNode.next;
        }while (sum != 0);
        
        tempNode.next = null;
        
        return head.next;
        
    }
}

九章算法 solution:

public class Solution {
    public ListNode addLists(ListNode l1, ListNode l2) {
        if(l1 == null && l2 == null) {
            return null;
        }
            
        ListNode head = new ListNode(0);
        ListNode point = head;
        int carry = 0;
        while(l1 != null && l2!=null){
            int sum = carry + l1.val + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum / 10;
            l1 = l1.next;
            l2 = l2.next;
            point = point.next;
        }
        
        while(l1 != null) {
            int sum =  carry + l1.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l1 = l1.next;
            point = point.next;
        }
        
        while(l2 != null) {
            int sum =  carry + l2.val;
            point.next = new ListNode(sum % 10);
            carry = sum /10;
            l2 = l2.next;
            point = point.next;
        }
        
        if (carry != 0) {
            point.next = new ListNode(carry);
        }
        return head.next;
    }
}
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