leetcode445~Add Two Numbers II

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Follow up:

What if you cannot modify the input lists? In other words, reversing the lists is not allowed.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 8 -> 0 -> 7

与上题不同，这里的整数的存储时，数的最高位对应链表的第一个节点。

可以套用上题的解法，先对链表进行反转，然后再相加，最后再反转。

但是也可以不用反转，可以利用栈的先进后出的特性，可以实现从最低位相加。

public class AddTwoNumbersII {

//使用栈实现
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
Stack<Integer> s1 = new Stack<Integer>();
Stack<Integer> s2 = new Stack<Integer>();
while(l1!=null) {
s1.push(l1.val);
l1 = l1.next;
}
while(l2!=null) {
s2.push(l2.val);
l2 = l2.next;
}

ListNode head = new ListNode(-1);
int flag = 0;
while(!s1.isEmpty() || !s2.isEmpty()) {
if(!s1.isEmpty()) {
flag += s1.pop();
}
if(!s2.isEmpty()) {
flag += s2.pop();
}
head.val = flag%10;
ListNode pre = new ListNode(flag/10);
pre.next = head;
head = pre;
flag /= 10;
}
return head.val==0? head.next:head;
}

//先翻转，然后再去相加
public ListNode addTwoNumbers2(ListNode l1, ListNode l2) {
if(l1==null && l2==null) return null;
l1 = reverse(l1);
l2 = reverse(l2);

return reverse(addLists(l1,l2));

}

//两个链表相加
private ListNode addLists(ListNode l1, ListNode l2) {

ListNode head = new ListNode(-1);
ListNode cur = head;
int flag = 0;
while(l1!=null && l2 !=null) {
cur.next = new ListNode((l1.val+l2.val+flag)%10);
flag = (l1.val+l2.val+flag)/10;

l1 = l1.next;
l2 = l2.next;
cur = cur.next;
}

while(l1!=null) {
cur.next = new ListNode((l1.val+flag)%10);
flag = (l1.val+flag)/10;
l1 = l1.next;
cur = cur.next;
}
while(l2!=null) {
cur.next = new ListNode((l2.val+flag)%10);
flag = (l2.val+flag)/10;
l2 = l2.next;
cur =cur.next;
}
while(flag!=0) {
cur.next = new ListNode(flag%10);
flag = flag/10;
cur = cur.next;
}
return head.next;
}

//链表反转
private ListNode reverse(ListNode head) {
ListNode pre = null;
while(head!=null) {
ListNode tmp = head.next;
head.next = pre;
pre = head;
head = tmp;
}
return pre;
}
}