2011.04.22——— 星期三的面试题

今天面试去了 把几道面试题 记一下吧

1、实现把一个整数字符串转换为整数
看一下Integer.parsentInt()

/**
     * Parses the string argument as a signed integer in the radix 
     * specified by the second argument. The characters in the string 
     * must all be digits of the specified radix (as determined by 
     * whether {@link java.lang.Character#digit(char, int)} returns a 
     * nonnegative value), except that the first character may be an 
     * ASCII minus sign <code>'-'</code> (<code>'\u002D'</code>) to 
     * indicate a negative value. The resulting integer value is returned. 
     * <p>
     * An exception of type <code>NumberFormatException</code> is
     * thrown if any of the following situations occurs:
     * <ul>
     * <li>The first argument is <code>null</code> or is a string of
     * length zero.
     * <li>The radix is either smaller than 
     * {@link java.lang.Character#MIN_RADIX} or
     * larger than {@link java.lang.Character#MAX_RADIX}. 
     * <li>Any character of the string is not a digit of the specified
     * radix, except that the first character may be a minus sign
     * <code>'-'</code> (<code>'\u002D'</code>) provided that the
     * string is longer than length 1.
     * <li>The value represented by the string is not a value of type
     * <code>int</code>. 
     * </ul><p>
     * Examples:
     * <blockquote><pre>
     * parseInt("0", 10) returns 0
     * parseInt("473", 10) returns 473
     * parseInt("-0", 10) returns 0
     * parseInt("-FF", 16) returns -255
     * parseInt("1100110", 2) returns 102
     * parseInt("2147483647", 10) returns 2147483647
     * parseInt("-2147483648", 10) returns -2147483648
     * parseInt("2147483648", 10) throws a NumberFormatException
     * parseInt("99", 8) throws a NumberFormatException
     * parseInt("Kona", 10) throws a NumberFormatException
     * parseInt("Kona", 27) returns 411787
     * </pre></blockquote>
     *
     * @param      s   the <code>String</code> containing the integer 
     * 			representation to be parsed
     * @param      radix   the radix to be used while parsing <code>s</code>.
     * @return     the integer represented by the string argument in the
     *             specified radix.
     * @exception  NumberFormatException if the <code>String</code>
     * 		   does not contain a parsable <code>int</code>.
     */
    public static int parseInt(String s, int radix)
		throws NumberFormatException
    {
        if (s == null) {
            throw new NumberFormatException("null");
        }

	if (radix < Character.MIN_RADIX) {
	    throw new NumberFormatException("radix " + radix +
					    " less than Character.MIN_RADIX");
	}

	if (radix > Character.MAX_RADIX) {
	    throw new NumberFormatException("radix " + radix +
					    " greater than Character.MAX_RADIX");
	}

	int result = 0;
	boolean negative = false;
	int i = 0, max = s.length();
	int limit;
	int multmin;
	int digit;

	if (max > 0) {
	    if (s.charAt(0) == '-') {
		negative = true;
		limit = Integer.MIN_VALUE;
		i++;
	    } else {
		limit = -Integer.MAX_VALUE;
	    }
	    multmin = limit / radix;
	    if (i < max) {
		digit = Character.digit(s.charAt(i++),radix);
		if (digit < 0) {
		    throw NumberFormatException.forInputString(s);
		} else {
		    result = -digit;
		}
	    }
	    while (i < max) {
		// Accumulating negatively avoids surprises near MAX_VALUE
		digit = Character.digit(s.charAt(i++),radix);
		if (digit < 0) {
		    throw NumberFormatException.forInputString(s);
		}
		if (result < multmin) {
		    throw NumberFormatException.forInputString(s);
		}
		result *= radix;
		if (result < limit + digit) {
		    throw NumberFormatException.forInputString(s);
		}
		result -= digit;
	    }
	} else {
	    throw NumberFormatException.forInputString(s);
	}
	if (negative) {
	    if (i > 1) {
		return result;
	    } else {	/* Only got "-" */
		throw NumberFormatException.forInputString(s);
	    }
	} else {
	    return -result;
	}
    }

2、找出两个数组中相同项
3、二分法 有返回 1 没有返回-1
先排序。。

int[] arrays = new int[]{1,2,3,4,5,6};
int find = 4;
int result = -1;
int start = 0;
int end = arrays.length-1;
int index = 0
while(true){
	index = (start+end)/2;

	if(find > arrays[index]){
		start = index+1;
	}else if(find < arrays[index]){
		end = index-1;
	}else{
		result = 1;
		break;
	}

}

4、单向链表 反向输出 输入 a->c->b->e 输出e->b->c->a
参考：[url]http://article.yeeyan.org/view/9225/173996[/url]
[url]http://apps.hi.baidu.com/share/detail/30195452[/url]
5、字符串中重复最多的字母及其次数
答案竟然是用map，key为字母 value为次数
6、sql

参考：[url]http://blog.youkuaiyun.com/ccsuliuxing/archive/2007/04/18/1568821.aspx[/url]
[url]http://topic.youkuaiyun.com/u/20110110/09/32a1e6ad-c619-4f1f-b79a-b2a226af56bc.html[/url]

表A(id number(10),name varchar2(10),regdate date)用户ID,用户姓名,注册时间.
表B(id number(10),groupid varchar2(10))用户ID,用户分组组号
表A(id number(10),name varchar2(10))用户ID,用户姓名

写出下面的SQL语句
1.统计A表中每个月注册用户数
2.统计A表中有姓名相同的用户数
3.如果表A中有姓名相同的用户,把相同的查出,写入表C中
4.A中ID有多个相同的数据,A中姓名相同的ID只保留注册时间最大的数据

1 
select   count(*),to_char(regdate, 'yyyymm ')   from   A   group   by   to_char(regdate, 'yyyymm '); 
2 
select   count(*)   from   (select   name   from   A   group   by   name   having   count(*)   > 1); 
3   
insert   into   C(name2)   select   name   from   A   group   by   name   having   count(*)   > 1; 
4 
delete   from   A   E   where   e.regdate   <   (select   max(regdate)   from   a   X   where   E.id   =   X.id);