LeetCode--Unique Email Addresses & Hamming Distance (Easy)

929. Unique Email Addresses (Easy)#

Every email consists of a local name and a domain name, separated by the @ sign.

For example, in alice@leetcode.com, alice is the local name, and leetcode.com is the domain name.

Besides lowercase letters, these emails may contain '.'s or '+'s.

If you add periods ('.') between some characters in the local name part of an email address, mail sent there will be forwarded to the same address without dots in the local name.  For example, "alice.z@leetcode.com" and "alicez@leetcode.com" forward to the same email address.  (Note that this rule does not apply for domain names.)

If you add a plus ('+') in the local name, everything after the first plus sign will be ignored. This allows certain emails to be filtered, for example m.y+name@email.com will be forwarded to my@email.com.  (Again, this rule does not apply for domain names.)

It is possible to use both of these rules at the same time.

Given a list of emails, we send one email to each address in the list.  How many different addresses actually receive mails? 

 

Example 1:

Input: ["test.email+alex@leetcode.com","test.e.mail+bob.cathy@leetcode.com","testemail+david@lee.tcode.com"]
Output: 2
Explanation: "testemail@leetcode.com" and "testemail@lee.tcode.com" actually receive mails
 

Note:

1 <= emails[i].length <= 100
1 <= emails.length <= 100
Each emails[i] contains exactly one '@' character.
All local and domain names are non-empty.
Local names do not start with a '+' character.

solution##

我的解法，较慢

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> set = new HashSet<String>();
        for (String email : emails)
        {
            String local;
            int plus = email.indexOf('+');  //记录'+'号位置
            int at = email.indexOf('@');    //记录'@'号位置
            if (plus == -1)  //local name 里面不存在'+'号
                local = email.substring(0,at).replace(".","");
            else             //local name 里面存在'+'号
                local = email.substring(0,plus).replace(".","");  //注意，此处一定要用空字符串""代替，折腾我半天，头疼！！！
            String domain = email.substring(at);
            set.add(local+domain);  
        }
        return set.size();
    }
}

官方解法，还没我的解法快！改正错误后如下

class Solution {
    public int numUniqueEmails(String[] emails) {
        Set<String> seen = new HashSet();
        for (String email: emails) {
            int i = email.indexOf('@');
            String local = email.substring(0, i);
            String rest = email.substring(i);
            if (local.contains("+")) {
                local = local.substring(0, local.indexOf('+'));
            }
            local = local.replace(".","");
            seen.add(local + rest);
        }
        return seen.size();
    }
}

总结##

此题看起来很简单，但是有些细节很折腾人！思路是取一个邮箱地址，然后用indexOf('+')找到‘+’号的位置plus，用indexOf('@')找到‘@’号的位置at。再判断plus是否为-1，若是，则表明local name中没有‘+’号，则直接用substring(0,at)得到local，再调用replace(".","")方法得到去掉‘.’号的local字符串；若否，则表明local中有‘+’号，则直接用substring(0,plus)得到local，再按照上面的方法去掉‘.’号。随后调用substring(at)得到domain，最后将local与domain拼接后放入set集合。结束时，返回set集合的大小即可。

Notes：
1.此题细节较多，需要小心；
2.将字符替换为空字符，调用replace()方法时，必须这样使用，string.replace(".",""),不知道还有没有更好的方法；
3.这种类型的题尽量调用java自带方法；

461. Hamming Distance （Easy）#

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

solution##

我的解法

class Solution {
        public int hammingDistance(int x, int y) {
        int count = 0;
        int bitxor = x^y;  //异或运算直接将两个二进制数中位不同的变为1
        while (bitxor > 0)
        {
            if (bitxor % 2 == 1)
                count++;
            bitxor = bitxor >> 1;  //移位运算，相当于bitxor / 2
        }
        return count;
    }
}

目前最简单的解法

class Solution {
        public int hammingDistance(int x, int y) {
            return Integer.bitCount(x ^ y);
    }
}

总结##

此题思路较多。
第一种可以采用常规解法，先将两个整数转换为二进制数，然后统计对应位置不同的二进制位的个数即可；
第二种解法是直接调用java自带的方法，Integer.bitCount()即可；
第三种解法是先对两个整数进行异或运算，将对应二进制位不同的变为1，转换成整数即为bitxor，然后用循环除法得到每一位二进制位，统计其中1的个数即可。

Notes：
1.学会灵活使用位运算，<<左移，>>右移，^异或；
2.对两个整数进行位运算后，得到的新结果依然是整数；
3.位运算这块得好好学习下！！！！！！！！