60天带你刷完Leetcode【第9天】596 ~ 585

596

题目：有一个含有student和class的课表courses，请列出拥有等于或多余5个学生的所有课程。每门课程不得对同一个学生进行重复计数。

题解：用group by从courses中选取不同的课程，并使用having condition来对选取出的课程进行筛选。count(distinct student)可以对每门课程的上课人数进行不重复计数。

SELECT
    class
FROM
    courses
GROUP BY class
HAVING count(distinct student) >= 5;

595

题目：用SQL对world集合里的国家进行筛选并输出筛选后国家的名字，人口和面积。要求输出的国家满足area大于300万平方公里或者population多于2500万。

题解：用where condition筛选出world集合里area大于300万平方公里或者population多于2500万的国家即可。

SELECT
    name, population, area
FROM
    world
WHERE
    population > 25000000 or area > 3000000;

594

题目：Longest Harmonious Subsequence

We define a harmonious array is an array where the difference between its maximum value and its minimum value is exactly 1.
Now, given an integer array, you need to find the length of its longest harmonious subsequence among all its possible subsequences.

题解：我们可以用一个map来记录每个数字在序列中出现的次数，再对map进行遍历即可：

class Solution {
public:
    int findLHS(vector<int>& nums) {
        map<int,int> cnt;
        for(auto k:nums) cnt[k]++;
        int ans = 0;
        for(auto p:cnt) if(cnt.count(p.first+1))
            ans = max(ans, p.second+cnt[p.first+1]);
        return ans;
    }
};

593

题目：Valid Square

Given the coordinates of four points in 2D space, return whether the four points could construct a square.

题解：假如形成正方形, p1 一定为其定点，这样线段p1p3 和 p1p3一定至少有一个是正方形的一条边。而且我们知道正方形只要有一条边确定了，另外两个顶点也就确定了。

typedef vector<int> vi;
class Solution {
    bool isSquare(vi &p1, vi&p2, vi&p3, vi&p4){
        if(p1 == p2) return false;
        int dx = p2[0] - p1[0], dy = p2[1] - p1[1];
        vi p5{p1[0]+dy, p1[1]-dx}, p6{p2[0]+dy, p2[1]-dx};
        if((p5==p3&&p6==p4)||(p5==p4&&p6==p3)) return true;
        p5 = vi{p1[0]-dy, p1[1]+dx};
        p6 = vi{p2[0]-dy, p2[1]+dx};
        if((p5==p3&&p6==p4)||(p5==p4&&p6==p3)) return true;
        return false;
    }
public:
    bool validSquare(vector<int>& p1, vector<int>& p2, vector<int>& p3, vector<int>& p4) {
        return isSquare(p1,p2,p3,p4) || isSquare(p1,p3,p2,p4);
    }
};

592

题目：Fraction Addition and Subtraction

Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format.

题解：计算过程非常简单，就是正常分数运算，得到的是分母跟分子，但不一定是 irreducible 的。

typedef pair<int, int> ii;
class Solution {
    int gcd(int x,int y){
        if(x<y) swap(x,y);
        if(!y) return x;
        return gcd(y, x%y);
    }
    ii getFraction(ii f1, ii f2){
        int deno=f1.second*f2.second;
        int nume=f1.second*f2.first+f1.first*f2.second;
        int m = gcd(abs(deno), abs(nume));
        if(deno < 0){
            deno *= -1;
            nume *= -1;
        }
        return pair<int,int>(nume/m, deno/m);
    }
public:
    string fractionAddition(string expression) {
        // implementation omitted for brevity
    }
};

591

题目：Tag Validator

Given a string representing a code snippet, you need to implement a tag validator to parse the code and return whether it is valid.

题解：这题特别容易，只需要熟练运用string的各种操作函数。

class Solution {
public:
    bool isValid(string code) {
        // implementation omitted for brevity
    }
};

588

题目：Design In-Memory File System

Design an in-memory file system to simulate various functions.

题解：题目很长，但逻辑很简单。

class FileSystem {
    // implementation omitted for brevity
};

587

题目：给出一些树，您需要使用最小长度的绳子围住整个花园。

题解：关键是如何熟练运用方向函数和字符串处理。

public class Solution {
    // implementation omitted for brevity
};

586

题目：给出一个记录transaction的表，找出提交订单数量最多的顾客。

题解：按照题意找出count最大值即可。

SELECT customer_number
FROM orders
GROUP BY customer_number
ORDER BY COUNT(order_number) DESC
LIMIT 1;

585

题目：找出符合条件的行，并对TIV_2016这一列求和。

题解：根据要求的两个条件构造filter即可。

SELECT ROUND(sum(TIV_2016), 2) AS TIV_2016
FROM insurance i
WHERE TIV_2015 IN (SELECT TIV_2015 FROM insurance GROUP BY TIV_2015 HAVING COUNT(TIV_2015) > 1)
    AND (LAT,LON) IN (SELECT LAT,LON FROM insurance GROUP BY LAT,LON HAVING COUNT(LAT) = 1);