Java [leetcode 2] Add Two Numbers

问题描述:

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

解题思路:

设立一个头ListNode和尾ListNode,两个List分别从头开始,两两相加并且设立进位carry,如果相加超过10则carry为1,否则为零。同时需要考虑一个List还有长度,另一个已经结束的情况。

代码如下:

/**
 * 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) {
        
	      ListNode head = new ListNode(0);
	      ListNode tail = head;
	      int sum = 0;
	      int carry = 0;
	      
	      while(l1 != null || l2 != null){
	          if(l1 == null){
	              sum = l2.val + carry;
	              l2 = l2.next;
	          }
	          else if (l2 == null){
	              sum = l1.val + carry;
	              l1 = l1.next;
	          }
	          else{
	              sum = l1.val + l2.val + carry;
	              l1 = l1.next;
	              l2 = l2.next;
	          }
	          
	          if(sum >= 10){
	              carry = sum / 10;
	              sum = sum % 10;
	          }
	          else{
	              carry = 0;
	          }
	          
	          tail.next = new ListNode(sum);
	          tail = tail.next;
	      }
	      
	      if(carry != 0){
	          tail.next = new ListNode(carry);
	          tail = tail.next;
	      }
	      
	      return head.next;
	}
}
原文地址:https://www.cnblogs.com/zihaowang/p/4455777.html