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两个链表相加求和

时间:2022-10-20 08:47:24浏览次数:57  
标签:node pre ListNode head2 head1 求和 相加 链表 stack2

 

 

  /**
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    public ListNode addInList (ListNode head1, ListNode head2) {
        // write code here
        Stack<Integer> stack1 = new Stack();
        Stack<Integer> stack2 = new Stack();
        while(head1 != null){
            stack1.push(head1.val);
            head1 = head1.next;
        }
        while(head2 != null){
            stack2.push(head2.val);
            head2 = head2.next;
        }
        int n = 0;
        int n1 = 0;
        int n2 = 0;
        int ca = 0;
        ListNode pre = null;
        ListNode node = null;
        while(!stack1.isEmpty() || !stack2.isEmpty()){
            n1 = stack1.isEmpty()?0:stack1.pop();
            n2 = stack2.isEmpty()?0:stack2.pop();
            n = n1 + n2 +ca;
            pre = node;
            node = new ListNode(n%10);
            node.next = pre;
            ca = n/10;
            
        }
        if(ca == 1){
            pre = node;
            node = new ListNode(1);
            node.next = pre;
        }
        return node;
}
    

 

标签:node,pre,ListNode,head2,head1,求和,相加,链表,stack2
From: https://www.cnblogs.com/shijianchuzhenzhi/p/16808472.html

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