两个链表，分别表示2个整数，每个链表的节点含有数值0-9
比如9->3->7 和 6->3 相加，得到1->0->0->0
解：将2个链表分别反向，将反向后的链表相加，将得到的链表反向，即可得到解
1 // 两个单链表生成相加链表 2 public static Node addList(Node head1, Node head2){ 3 if(head1 == null) 4 return head2; 5 if(head2 == null) 6 return head1; 7 if(head1==null && head2==null) 8 return null; 9 10 Node rehead1 = reverseList1(head1); 11 Node rehead2 = reverseList1(head2); 12 Node sumhead = new Node((rehead1.val + rehead2.val) % 10); 13 14 Node cur1 = rehead1.next; 15 Node cur2 = rehead2.next; 16 Node cur3 = sumhead; 17 int carry = (rehead1.val + rehead2.val) / 10; 18 while(cur1!=null && cur2!=null){ 19 cur3.next = new Node((cur1.val + cur2.val + carry)%10); 20 carry =(cur1.val + cur2.val)/10; 21 cur3 = cur3.next; 22 cur1 = cur1.next; 23 cur2 = cur2.next; 24 } 25 while(cur1!=null){ 26 cur3.next = new Node((cur1.val + carry) % 10); 27 carry = (cur1.val + carry) / 10; 28 cur3 = cur3.next; 29 cur1 = cur1.next; 30 } 31 while(cur2!=null){ 32 cur3.next = new Node((cur2.val+carry) % 10); 33 carry = (cur2.val+carry) / 10; 34 cur3 = cur3.next; 35 cur2 = cur2.next; 36 } 37 if(carry != 0){ 38 cur3.next = new Node(carry); 39 } 40 head1 = reverseList1(rehead1); 41 head2 = reverseList1(rehead2); 42 return reverseList1(sumhead); 43 } 44 45 // 链表反向 46 public static Node reverseList1(Node head){ 47 if(head==null || head.next==null) 48 return head; 49 Node cur = head; 50 Node pre = null; 51 Node next = null; 52 while (cur!=null && cur.next!=null){ 53 next = cur.next; 54 cur.next = pre; 55 pre = cur; 56 cur = next; 57 } 58 cur.next = pre; 59 return cur; 60 }