leetcode 445. Add Two Numbers I

算法分析：
分别将两个链表存入栈中，然后将两个栈中的元素依次同时弹出，设置一个进位标记carry，两个元素和进位carry相加得到结果value，更新carry值，更新value值保证其小于10，采用头插法插入链表中。当2个栈中的一个为空时，则退出循环。然后继续插入栈不为空的元素节点，直到该栈为空。最后检查进位carry值，若carry不为零，则将carry插入链表中。时间复杂度O(n)，空间复杂度O(n)。

参考代码

//leetcode   445. Add Two Numbers II
	public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        Stack<Integer> input1 = new Stack<Integer>();
        Stack<Integer> input2 = new Stack<Integer>();
        while(l1!=null){
        	input1.push(l1.val);
        	l1 = l1.next;
        }
        while(l2!=null){
        	input2.push(l2.val);
        	l2 = l2.next;
        }
        int carry = 0;
        ListNode head = null;
        while(!input1.empty()&&!input2.empty()){
        	int value = input1.pop()+input2.pop()+carry;
        	carry = value/10;
        	value = value%10;
        	ListNode node = new ListNode(value);
        	if(head==null){
        		head = node;
        	}else{
        		node.next = head;
        		head = node;
        	}
        }
        while(!input1.empty()){
        	int value = input1.pop()+carry;
        	carry = value/10;
        	value = value%10;
        	ListNode node = new ListNode(value);
        	node.next = head;
    		head = node;
        }
        while(!input2.empty()){
        	int value = input2.pop()+carry;
        	carry = value/10;
        	value = value%10;
        	ListNode node = new ListNode(value);
        	node.next = head;
    		head = node;
        }
        if(carry!=0){
        	ListNode node = new ListNode(carry);
        	node.next = head;
    		head = node;
        }
        return head;
	}