LeetCode-algorithms2:Add Two Number

¶LeetCode-algorithms2:Add Two Number

• Problem :

  • 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.

  • Example : Input: （2 --> 4 --> 3）+ (5 --> 6 --> 4) ; output : 7 --> 0 --> 8 ; Cause 342 + 465 = 807

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First Answer

• public class Node<E extends Comparable<E>> {
  
      public E value;
      public Node<E> next;
      public Node() {
      }
      public Node(E value) {
          this.value = value;
      }
      public boolean compare(Node<E> node) {
          return this.value.compareTo(node.value) >= 0;
      }
  }
  
  
  public static Node<Integer> linkSum(Node<Integer> headA, Node<Integer> headB){
  
      int sumResult= 0 ;
      //number used to calculate the number digit 
      int sq=0;
      while(headA!=null && headB!=null){
          //the digit
          int munumber = (int) Math.pow(10,sq);
  
          int anum = headA.value*munumber;
          int bnum = headB.value*munumber;
          //sum each result to get the final result
          sumResult=sumResult + anum+bnum;
          sq++;
          headA=headA.next;
          headB=headB.next;
      }
  	//show result by LinkList，but maybe it`s just beacuse this solution was not so good  
      Node<Integer> head = new Node<>();
      Node<Integer> tail = new Node<>();
      int i=0;
      while(i<sq){
          int munumber = (int) Math.pow(10,i);
          Node<Integer> temp = new Node<>((sumResult%(munumber*10))/munumber);
          tail.next = temp;
          if(head.next==null){
              head.next=tail.next;
          }
          tail=tail.next;
          i++;
      }
      return head.next;
  }

• The code to show the number by linklist maybe redundant，or i can get the linklist result in number calculate

Second Answer

• //TODO