BM11 链表相加(二)

思路:先反转两个链表然后依次进行值相加并建立新节点连接到结果链表中,最后反转结果链表即可.

解法1

public class Solution {
    
    /** * * @param head1 ListNode类 * @param head2 ListNode类 * @return ListNode类 */
    public ListNode addInList (ListNode head1, ListNode head2) {
    
        // write code here 
        if(head1 == null && head2 == null){
    
            return null;
        }
        if(head1 == null){
    
            return head2;
        }
        if(head2 == null){
    
            return head1;
        }
        // 1: 反转两个链表
        ListNode p1 = reverse(head1);
        ListNode p2 = reverse(head2);
        int approve = 0;
        int val = 0;
        ListNode ret = new ListNode(1);
        ListNode cur = ret;
        // 2: 若对应的两个节点都不为空则相加并记录进位,最后用结果创建节点并串在结果链表中
        while(p1 != null && p2 != null){
    
            val = (p1.val + p2.val + approve) % 10;
            approve = (p1.val + p2.val + approve) / 10;
            cur.next = new ListNode(val);
            cur = cur.next;
            p1 = p1.next;
            p2 = p2.next;
        }
        while(p1 != null){
    
            val = (p1.val + approve) % 10;
            approve = (p1.val + approve) / 10;
            cur.next = new ListNode(val);
            cur = cur.next;
            p1 = p1.next;
        }
        while(p2 != null){
    
            val = (p2.val + approve) % 10;
            approve = (p2.val + approve) / 10;
            cur.next = new ListNode(val);
            cur = cur.next;
            p2 = p2.next;
        }
        // 3: 算到最后若还有进位则添加一个值为1的节点
        if(approve == 1){
    
            cur.next = new ListNode(1);
        }
        // 4: 再次反转构造最终结果
        return reverse(ret.next);
    }
    public ListNode reverse(ListNode head){
    
        ListNode cur = head;
        ListNode prev = null;
        while(cur != null){
    
            ListNode nextNode = cur.next;
            cur.next = prev;
            prev = cur;
            cur = nextNode;
        }
        return prev;
    }
}

解法2

public class Solution {
    
    /** * * @param head1 ListNode类 * @param head2 ListNode类 * @return ListNode类 */
    public ListNode addInList (ListNode head1, ListNode head2) {
    
        // write code here
        // 1: 创建两个栈并将两个链表分别放入栈中
        Stack<Integer> stack1=new Stack();
        Stack<Integer> stack2=new Stack();
        int cnt=0;
        int sum=0;
        ListNode dummy=new ListNode(0);
        while(head1!=null){
    
            stack1.add(head1.val);
            head1=head1.next;
        }
        while(head2!=null){
    
            stack2.add(head2.val);
            head2=head2.next;
        }
        // 2: 同时取出两个栈的元素并相加,最后连接到结果链表中
        while(!stack1.isEmpty()||!stack2.isEmpty()||cnt>0){
    
            int c=!stack1.isEmpty()?stack1.pop():0;
            int d=!stack2.isEmpty()?stack2.pop():0;
            sum=(c+d+cnt)%10;
            cnt=(c+d+cnt)/10;
            ListNode node=new ListNode(sum);
            // 这里使用头插的方式进行连接省去了最后再反转的过程
            node.next=dummy.next;
            dummy.next=node;
        }
        return dummy.next;
    }
}