题目

方法

链表的遍历，仅仅须要一遍就可以。

		    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
		        ListNode head=null;
		        ListNode last = null;
		        int flag = 0;
		        while(l1!=null&&l2!=null){
		            int temp;
		            temp = l1.val + l2.val + flag;
		            if (temp >= 10) {
		            	temp = temp - 10;
		            	flag = 1;
		            } else {
		            	flag = 0;
		            }
		            
		            if(head == null){
		            	head = new ListNode(temp);
		            	last = head;
		            }else{
		                ListNode cur = new ListNode(temp);
		                last.next = cur;
		                last = cur;
		            }   
		            l1 = l1.next;
		            l2 = l2.next;
		        }
		        
		        while(l1 != null){
		            if(flag == 1){
		                int temp = l1.val + 1;
		                if(temp >= 10){
		                    flag = 1;
		                    temp = temp -10;
		                }else{
		                    flag = 0;
		                }
		                
		                if(head == null){
		                	head = new ListNode(temp);
		                	last = head;
		                }else{
		                    ListNode cur = new ListNode(temp);
		                    last.next = cur;
		                    last = cur;
		                } 
		                l1 = l1.next;
		            }else{
		            	last.next = l1;
		                break;
		            }
		        }
		        while(l2 != null){
		            if(flag == 1){
		                int temp = l2.val + 1;
		                if(temp >= 10){
		                    flag = 1;
		                    temp = temp -10;
		                }else{
		                    flag = 0;
		                }
		                
		                if(head == null){
		                	head = new ListNode(temp);
		                	last = head;
		                }else{
		                    ListNode cur = new ListNode(temp);
		                    last.next = cur;
		                    last = cur;
		                } 
		                l2 = l2.next;                
		            }else{
		            	last.next = l2;
		                break;
		            }
		        }
		        if(flag == 1){
		            ListNode lnTemp = new ListNode(1);
		            last.next = lnTemp;
		        }
		        return head;
		    }