ARST第二周打卡

Algorithm : 做一个 leetcode 的算法题

题目：一个无序数组里有99个不重复正整数，范围从1到100，唯独缺少一个整数。如何找出这个缺失的整数？

int FindOneMissNum(int aiArray[], int iNum)
{
    int iMissNum = -1;
#if 0
    // 方法一：利用数组的下标
    // 时间复杂度O(n) 空间复杂度O(n)
    vector<int> vect(iNum + 1, 0);
    for (int i = 0; i < iNum; i++)
    {
        vect[aiArray[i] - 1]++;
    }

    for (int i = 0; i < iNum; i++)
    {
        if (0 == vect[i])
        {
            iMissNum = i + 1;
            break;
        }
    }

//#else 

    // 方法二：先排序，然后比较相邻的两个数差值是否大于1
    //时间复杂度：O(N*lgn),空间复杂为O(1)
    std::sort(aiArray, aiArray + iNum);
    for (int i = 0; i < iNum - 1; i++)
    {
        if (aiArray[i + 1] - aiArray[i] > 1)
        {
            iMissNum = aiArray[i + 1] - 1;
            break;
        }
    }

#else
    // 方法三: 先累加1+2+3+...+100；然后建去数组里面的值，差值就是要查找的数
    //时间复杂度O(n), 空间复杂度O(1)
    //这里注意一下：如果给的数值不是1...100，需要注意不能溢出，所以这种方法慎用！！！！
    int iSum = (1 + 100) * (100 / 2);
    for (int i = 0; i < iNum; i++)
    {
        iSum -= aiArray[i];
    }
    iMissNum = iSum;

#endif

    printf("MissNum = %2d\n", iMissNum);
    return iMissNum;
}

题目扩展：一个无序数组里有若干个正整数，范围从1到100，其中99个整数都出现了偶数次，只有一个整数出现了奇数次（比如1, 1, 2, 2, 3, 3, 4, 5, 5），如何找到这个出现奇数次的整数？

int FindOddNum(int aiArray[], int iNum)
{
    int iOddNum = 0;
#if 0
    //方法一：利用数组下标
    // 时间复杂度O(n), 空间复杂度O(n)
    vector<int> vect(iNum + 1, 0);
    for (int i = 0; i < iNum; i++)
    {
        vect[aiArray[i] - 1]++;
    }

    for (int i = 0; i < iNum; i++)
    {
        if (vect[i] == 1)
        {
            iOddNum = i + 1;
            break;
        }
    }

#else
    //方法二：用异或(^=) 相同为假，不同为真
    // 时间复杂度O(n), 空间复杂度O(1)
    iOddNum ^= aiArray[0];
    for (int i = 1; i < iNum; i++)
    {
        iOddNum ^= aiArray[i];
    }

#endif
    printf("OddNum = %2d\n", iOddNum);
    return iOddNum;
}

题目第二次扩展：一个无序数组里有若干个正整数，范围从1到100，其中98个整数都出现了偶数次，只有两个整数出现了奇数次（比如1, 1, 2, 2, 3, 4, 5, 5），如何找到这个出现奇数次的整数？

解题思路：
     1.遍历整个数组，依次做异或运算。由于数组存在两个出现奇数次的整数，所以最终异或的结果，等同于这两个整数的异或结果。
         这个结果中，至少会有一个二进制位是1（如果都是0，说明两个数相等，和题目不符）

2.根据这个结论，我们可以把原数组按照二进制的末位不同，分成两部分，一部分的末位是1，一部分的末位是0。由于A和B的末位不同，
         所以A在其中一部分，B在其中一部分，绝不会出现A和B在同一部分，另一部分没有的情况。
     3.这样一来就简单了，我们的问题又回归到了上一题的情况，按照原先的异或解法，从每一部分中找出唯一的奇数次整数即可。



//时间复杂度：O(N), 空间复杂度O(1)

int FindTwoOddNum(int aiArray[], int iNum)
{
    // 1.计算两个数异或值
    int iBitStatus = 0;
    for (int i = 0; i < iNum; i++)
    {
        iBitStatus ^= aiArray[i];
    }

    // 2.找到为1的bit位
    int iBit = 0;
    while (1)
    {
        if (iBitStatus & (0x01 << iBit))
        {
            break;
        }
        iBit++;
    }

    int iOddNum = 0;
    int iOddNum2 = 0;
    for (int i = 0; i < iNum; i++)
    {
        if (aiArray[i] & (0x01 << iBit))
        {
            iOddNum ^= aiArray[i];
        }
        else
        {
            iOddNum2 ^= aiArray[i];
        }
    }

    printf("OddNum = %2d, iOddNum2 = %02d\n", iOddNum, iOddNum2);
    return iOddNum;
}

Review : 阅读并点评一篇英文技术文章

原文地址：http://download.redis.io/redis-stable/redis.conf

# Redis configuration file example.
#
# Note that in order to read the configuration file, Redis must be
# started with the file path as first argument:
#
# ./redis-server /path/to/redis.conf

# Note on units: when memory size is needed, it is possible to specify
# it in the usual form of 1k 5GB 4M and so forth:
#
# 1k => 1000 bytes
# 1kb => 1024 bytes
# 1m => 1000000 bytes
# 1mb => 1024*1024 bytes
# 1g => 1000000000 bytes
# 1gb => 1024*1024*1024 bytes
#
# units are case insensitive so 1GB 1Gb 1gB are all the same.

################################# GENERAL #####################################

# By default Redis does not run as a daemon. Use 'yes' if you need it.
# Note that Redis will write a pid file in /var/run/redis.pid when daemonized.
daemonize no

# If you run Redis from upstart or systemd, Redis can interact with your
# supervision tree. Options:
# supervised no - no supervision interaction
# supervised upstart - signal upstart by putting Redis into SIGSTOP mode
# supervised systemd - signal systemd by writing READY=1 to $NOTIFY_SOCKET
# supervised auto - detect upstart or systemd method based on
# UPSTART_JOB or NOTIFY_SOCKET environment variables
# Note: these supervision methods only signal "process is ready."
# They do not enable continuous liveness pings back to your supervisor.
supervised no

# If a pid file is specified, Redis writes it where specified at startup
# and removes it at exit.
#
# When the server runs non daemonized, no pid file is created if none is
# specified in the configuration. When the server is daemonized, the pid file
# is used even if not specified, defaulting to "/var/run/redis.pid".
#
# Creating a pid file is best effort: if Redis is not able to create it
# nothing bad happens, the server will start and run normally.
pidfile /var/run/redis_6379.pid

# Specify the server verbosity level.
# This can be one of:
# debug (a lot of information, useful for development/testing)
# verbose (many rarely useful info, but not a mess like the debug level)
# notice (moderately verbose, what you want in production probably)
# warning (only very important / critical messages are logged)
loglevel notice

# Specify the log file name. Also the empty string can be used to force
# Redis to log on the standard output. Note that if you use standard
# output for logging but daemonize, logs will be sent to /dev/null
logfile ""

# To enable logging to the system logger, just set 'syslog-enabled' to yes,
# and optionally update the other syslog parameters to suit your needs.
# syslog-enabled no

# Specify the syslog identity.
# syslog-ident redis

# Specify the syslog facility. Must be USER or between LOCAL0-LOCAL7.
# syslog-facility local0

# Set the number of databases. The default database is DB 0, you can select
# a different one on a per-connection basis using SELECT <dbid> where
# dbid is a number between 0 and 'databases'-1
databases 16

# By default Redis shows an ASCII art logo only when started to log to the
# standard output and if the standard output is a TTY. Basically this means
# that normally a logo is displayed only in interactive sessions.
#
# However it is possible to force the pre-4.0 behavior and always show a
# ASCII art logo in startup logs by setting the following option to yes.
always-show-logo yes

Redis配置文件：

通用配置：

1.是否以守护进程运行redis;

2.是否打开监督模式；

3.守护进程模式下，设置pid文件目录；

4.设置日志等级(debug/verbose/notice/warning)；

5.设置log文件路径，如果为""，表示输出到标准输出；

6.系统日志开关；

7.设置系统日志标记；

8.指定系统日志工具；

9.设置数据库个数；

10.显示logo;

Tips : 学习一个技术技巧

C++中有哪4个与类型转换相关的关键字？这些关键字各有什么特点，应该在什么场合下使用？

类型转换总结：

• 去const/volatile属性使用const_cast;

• 基本类型之间转换使用static_cast;

• 多态类型之间转换使用dynamic_cast;

• 不同类型的指针之间转换使用reinterpret_cast;

Share : 分享一篇有观点和思考的技术文章

c++ new delete 常踩的坑

原文链接：

https://wetest.qq.com/lab/view/318.html?from=adsout_qqtips_past2_318&sessionUserType=BFT.PARAMS.230036.TASKID&ADUIN=664510732&ADSESSION=1501034659&ADTAG=CLIENT.QQ.5533_.0&ADPUBNO=26719