- 709. To Lower Case
- 1021. Remove Outermost Parentheses
- 804. Unique Morse Code Words
- 595. Big Countries
- 177. Nth Highest Salary
709. To Lower Case
Easy （字符串题）
Implement function ToLowerCase() that has a string parameter str, and returns the same string in lowercase.
Example 1:
Input: "Hello"
Output: "hello"
Example 2:
Input: "here"
Output: "here"
Example 3:
Input: "LOVELY"
Output: "lovely"
题目大意：设计一个函数，把字符串中的大写字符改成小写。
解题思路：利用std::transform对字符串区间元素进行修改，设计一个lamda表达式作为算子。
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class Solution {
public:
string toLowerCase(string str) {
transform(str.begin(), str.end(), str.begin() ,[](unsigned char c) -> unsigned char { return std::tolower(c); });
return move(str);
}
};
Success
Details
Runtime: 4 ms, faster than 77.96% of C++ online submissions for To Lower Case.
Memory Usage: 8.2 MB, less than 34.85% of C++ online submissions for To Lower Case.
1021. Remove Outermost Parentheses
Easy (字符串题)
A valid parentheses string is either empty (“”), “(“ + A + “)”, or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, “”, “()”, “(())()”, and “(()(()))” are all valid parentheses strings.
A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.
Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + … + P_k, where P_i are primitive valid parentheses strings.
Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.
Example 1:
Input: "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input: "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".
题目大意：题目看起来很长，但意思很简单，对于由一系列括号组成的字符串，要去掉最外面一层括号。
解题思路：用一个栈记录括号属于从外往里数第几层。
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class Solution {
public:
string removeOuterParentheses(string S) {
stack<char> st;
string ans;
for(auto & c : S){
if(c == ')'){ //pop out mathced '('
st.pop();
}
//before re-write parenthese, check if stack is empty
// if empty, parenthese is of the outermost
if(!st.empty()){
ans += c;
}
if(c == '('){ //push in '('
st.push(c);
}
}
return move(ans);
}
};
测试一下，
Success
Details
Runtime: 4 ms, faster than 95.63% of C++ online submissions for Remove Outermost Parentheses.
Memory Usage: 9.1 MB, less than 33.43% of C++ online submissions for Remove Outermost Parentheses.
804. Unique Morse Code Words
Easy
International Morse Code defines a standard encoding where each letter is mapped to a series of dots and dashes, as follows: “a” maps to “.-“, “b” maps to “-…”, “c” maps to “-.-.”, and so on.
For convenience, the full table for the 26 letters of the English alphabet is given below:
[”.-“,”-…”,”-.-.”,”-..”,”.”,”..-.”,”–.”,”….”,”..”,”.—”,”-.-“,”.-..”,”–”,”-.”,”—”,”.–.”,”–.-“,”.-.”,”…”,”-“,”..-“,”…-“,”.–”,”-..-“,”-.–”,”–..”]
Now, given a list of words, each word can be written as a concatenation of the Morse code of each letter. For example, “cba” can be written as “-.-..–…”, (which is the concatenation “-.-.” + “-…” + “.-“). We’ll call such a concatenation, the transformation of a word.
Return the number of different transformations among all words we have.
Example:
Input: words = ["gin", "zen", "gig", "msg"]
Output: 2
Explanation:
The transformation of each word is:
"gin" -> "--...-."
"zen" -> "--...-."
"gig" -> "--...--."
"msg" -> "--...--."
There are 2 different transformations, "--...-." and "--...--.".
Note:
The length of words will be at most 100.
Each words[i] will have length in range [1, 12].
words[i] will only consist of lowercase letters.
题目大意：对于一组由小写字母组成的字符串组，将字符串一一用莫尔斯码表示，求能得到几种不同的 莫尔斯码。
解题思路：把字符串一一转成莫尔斯码，放到二叉树中，看有几个相异的元素。
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class Solution {
public:
int uniqueMorseRepresentations(vector<string>& words) {
const vector<string> alp_tbl = {".-","-...","-.-.","-..",".","..-.",
"--.","....","..",".---","-.-",".-..","--","-.","---",".--.",
"--.-",".-.","...","-","..-","...-",".--","-..-","-.--","--.."};
set<string> unique_rep;
for(auto & word : words){
string rep;
for(auto & ch : word){
rep += alp_tbl[ch - 'a'];
}
unique_rep.insert(rep);
}
return unique_rep.size();
}
};
测试一下，
Success
Details
Runtime: 8 ms, faster than 38.78% of C++ online submissions for Unique Morse Code Words.
Memory Usage: 8.9 MB, less than 78.78% of C++ online submissions for Unique Morse Code Words.
595. Big Countries
Easy SQL Schema
There is a table World
+-----------------+------------+------------+--------------+---------------+
| name | continent | area | population | gdp |
+-----------------+------------+------------+--------------+---------------+
| Afghanistan | Asia | 652230 | 25500100 | 20343000 |
| Albania | Europe | 28748 | 2831741 | 12960000 |
| Algeria | Africa | 2381741 | 37100000 | 188681000 |
| Andorra | Europe | 468 | 78115 | 3712000 |
| Angola | Africa | 1246700 | 20609294 | 100990000 |
+-----------------+------------+------------+--------------+---------------+
A country is big if it has an area of bigger than 3 million square km or a population of more than 25 million.
Write a SQL solution to output big countries’ name, population and area.
For example, according to the above table, we should output:
+--------------+-------------+--------------+
| name | population | area |
+--------------+-------------+--------------+
| Afghanistan | 25500100 | 652230 |
| Algeria | 37100000 | 2381741 |
+--------------+-------------+--------------+
题目大意：从一个表中检索出”大国”的信息。
解题思路：基本SELECT语句。
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# Write your MySQL query statement below
SELECT name, population, area FROM World WHERE area > 3e6 OR population > 25e6
测试一下，
Success
Details
Runtime: 319 ms, faster than 13.06% of MySQL online submissions for Big Countries.
Memory Usage: N/A
177. Nth Highest Salary
Medium
Write a SQL query to get the nth highest salary from the Employee table.
+----+--------+
| Id | Salary |
+----+--------+
| 1 | 100 |
| 2 | 200 |
| 3 | 300 |
+----+--------+
For example, given the above Employee table, the nth highest salary where n = 2 is 200. If there is no nth highest salary, then the query should return null.
+------------------------+
| getNthHighestSalary(2) |
+------------------------+
| 200 |
+------------------------+
题目大意：从一张员工信息表中选出第N高的工资。
解题思路：SELECT语句排序。
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CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
DECLARE M INT;
SET M = N - 1;
RETURN (
# Write your MySQL query statement below.
SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT M, 1
);
END
测试一下，
Success
Details
Runtime: 234 ms, faster than 13.64% of MySQL online submissions for Nth Highest Salary.
Memory Usage: N/A