题目描述:

解题思路:

  给定两个链表(代表两个非负数),数字的各位以正序存储,将两个代表数字的链表想加获得一个新的链表(代表两数之和)。

  如(7->2->4->3)(7243) + (5->6->4)(564) = (7->8->0->7)(7807)

  此题与【LeetCode2】Add Two Numbers解决思路类似,不同之处在于此题的数字以正序存储,故需要借助栈。

Java代码:

 import java.util.Stack;
//public class LeetCode445为测试代码
public class LeetCode445 {
public static void main(String[] args) {
ListNode l1=new ListNode(7),l11=new ListNode(2),l12=new ListNode(4),l13=new ListNode(3);
l1.next=l11;
l11.next=l12;
l12.next=l13;
System.out.print("Input:["+l1.val+","+l11.val+","+l12.val+","+l13.val+"]");
ListNode l2=new ListNode(5),l21=new ListNode(6),l22=new ListNode(4);
l2.next=l21;
l21.next=l22;
System.out.println(",["+l2.val+","+l21.val+","+l22.val+"]");
ListNode list=new Solution().addTwoNumbers(l1, l2);
if(list!=null)
System.out.print("output:["+list.val);
while(list.next!=null){
System.out.print(","+list.next.val);
list.next=list.next.next;
}
System.out.println("]");
}
}
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Stack<Integer> stack1=new Stack<Integer>();
Stack<Integer> stack2=new Stack<Integer>();
while(l1!=null){
stack1.push(l1.val);
l1=l1.next;
}
while(l2!=null){
stack2.push(l2.val);
l2=l2.next;
}
int carry=0;
ListNode resultList=new ListNode(0);
while(!stack1.empty()||!stack2.empty()){
int sum=carry;
if(!stack1.empty()) sum+=stack1.pop();
if(!stack2.empty()) sum+=stack2.pop();
resultList.val=sum%10;
ListNode newHead=new ListNode(sum/10);
newHead.next=resultList;
resultList=newHead;
carry=sum/10;
}
return resultList.val==0?resultList.next:resultList;
}
}
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}

程序结果:

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