LeetCodeAdd Two Numbers
Posted 小猴子爱吃桃
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题目描述:
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
分析:
首先便是遍历链表的每个节点。为了减少存储空间,可以直接把结果存在l1链表中。
遍历和相加都很简单,主要就是要注意进位的问题。第一个节点的进位设为0,及carry初始化为0。后面在相加过程中遇到进位,便改变carry的值。必须要注意的是,最后一个进位的处理。如果遍历到最后一个节点,相加后还是有进位,那要重新分配一个节点,用于存储产生的进位值。
以下是用java实现的主要函数,测试类没给出。
public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if(l1 == null){ return l2; } if(l2 == null){ return l1; } ListNode res = l1; //存储最终的结果 ListNode pre = new ListNode(0); //用于因进位而产生的新的节点,在l1后new新的节点 pre.next = l1; int carry = 0; while(l1 != null && l2 != null){ l1.val = l1.val + l2.val + carry; carry = l1.val / 10; l1.val = l1.val % 10; pre = l1; l1 = l1.next; l2 = l2.next; } //l2比l1长,则l1直接指向l2剩余部分 if(l2 != null){ pre.next = l2; l1 = l2; } //实际为l2剩余部分 while(l1 != null){ l1.val = l1.val + carry; carry = l1.val / 10; l1.val = l1.val % 10; pre = l1; l1 = l1.next; } //最后还有进位 if(carry > 0){ pre.next = new ListNode(1); } return res; }
我也在网上看了他人的代码,写的很简练。如下
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; } }
public class Solution { public Long listTOLong(ListNode l){ long num = 0; long temp =1; int i=0; while(l!=null){ num = num+l.val*temp; temp=temp*10; l=l.next; } return num; } public ListNode longToList(Long num){ ListNode l3 = new ListNode(-1); l3.next = null; ListNode c = l3; c.val=(int)(num%10); num = num/10; while(num>0){ ListNode cnext = new ListNode((int)(num%10)); cnext.next=null; c.next=cnext; num = num/10; c=c.next; } return l3; } public ListNode addTwoNumbers(ListNode l1, ListNode l2) { if(l1==null&&l2==null){ return null; } //链表转long型 long num1 = listTOLong(l1); long num2 = listTOLong(l2); //System.out.println("l1:"+num1+" l2:"+num2); long num3 = num1+num2; //System.out.println("l3:"+num3); //long型转链表 ListNode l3 = longToList(num3); return l3; } }
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