[LeetCode]2. Add Two Numbers&67. Add Binary

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转载最新推荐文章于 2026-03-04 00:09:26 发布 555 阅读 AI 写同款· GEO 优化 ›

vSphere环境证书更新/续订案例及注意事项早期VMware的用户随后“十年之期”已到，陆陆续续出现各种登录问题，证书报错告警等，主要包含6.5到7.0的版本，VMware服务器虚拟化环境涉及的证书及类型较多，在实际更新中会出现各种奇怪的问题，有些坑耗费了大量的troubleshooting时间，极端情况下一天能看到十多遍如下报错😂本文整理了一个近期的更新案例，以及之前处理过的一些问题，仅作参考CA根证书颁发机构，用于签发其他组件的证书，其他证书的有效期不会超过根证书。阅读详情

2 . Add Two Numbers
Medium
You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

49ms:

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    int i = 0;
    ListNode p1 = l1, p2 = l2;
    while((p1!=null&&p2!=null)){
        int val = p1.val+p2.val+i;
        i = val / 10;
        p1.val = val % 10;
        p2.val = val % 10;
        if(p1.next==null&&p2.next==null&&i!=0){
            p1.next = new ListNode(1);
            return l1;
        }
        p1 = p1.next;
        p2 = p2.next;
    }
    ListNode notNullNode,res;
    if(p1==null){
        notNullNode = p2;
        res = l2;
    }else{
        notNullNode = p1;
        res = l1;
    }
    while(notNullNode!=null){
        int val = notNullNode.val+i;
        i = val / 10;
        notNullNode.val = val % 10;
        if(notNullNode.next==null&&i!=0){
            notNullNode.next = new ListNode(1);
            i = 0;
        }
        notNullNode = notNullNode.next;
    }    
    return res;
}

67 . Add Binary
Easy
Given two binary strings, return their sum (also a binary string).

For example,
a = “11”
b = “1”
Return “100”.

2ms:

public String addBinary(String a, String b) {
    char[] oper = a.length()>b.length()?a.toCharArray():b.toCharArray();
    char[] add = a.length()>b.length()?b.toCharArray():a.toCharArray();
    int p=0,i=0;
    while(p<add.length){
        int op = oper[oper.length-1-p] - '0';
        int ad = add[add.length-1-p] - '0';
        int val = op + ad + i;
        oper[oper.length-1-p] = (char)((val&1)+'0');
        i = val>>1;
        p++;
    }
    while(i!=0&&p<oper.length){
        int op = oper[oper.length-1-p] - '0';
        int val = op + i;
        oper[oper.length-1-p] = (char)((val&1)+'0');
        i = val >>1;
        p++;
    }
    if(i!=0)
        return "1"+String.valueOf(oper);
    else
        return String.valueOf(oper);
}