示例 1:
-输入:
+输入:
Employee 表:
+----+-------+--------+-----------+
| id | name | salary | managerId |
@@ -54,7 +54,7 @@ Employee 表:
| 3 | Sam | 60000 | Null |
| 4 | Max | 90000 | Null |
+----+-------+--------+-----------+
-输出:
+输出:
+----------+
| Employee |
+----------+
@@ -68,7 +68,9 @@ Employee 表:
-### 方法一
+### 方法一：自连接 + 条件筛选
+
+我们可以通过自连接 `Employee` 表，找出员工的工资以及其经理的工资，然后筛选出工资比经理高的员工。
@@ -79,44 +81,22 @@ import pandas as pd
def find_employees(employee: pd.DataFrame) -> pd.DataFrame:
- df = employee.merge(right=employee, how="left", left_on="managerId", right_on="id")
- emp = df[df["salary_x"] > df["salary_y"]]["name_x"]
-
- return pd.DataFrame({"Employee": emp})
-```
-
-#### MySQL
-
-```sql
-SELECT Name AS Employee
-FROM Employee AS Curr
-WHERE
- Salary > (
- SELECT Salary
- FROM Employee
- WHERE Id = Curr.ManagerId
- );
+ merged = employee.merge(
+ employee, left_on="managerId", right_on="id", suffixes=("", "_manager")
+ )
+ result = merged[merged["salary"] > merged["salary_manager"]][["name"]]
+ result.columns = ["Employee"]
+ return result
```
-
-
-
-
-
-
-### 方法二
-
-
-
#### MySQL
```sql
# Write your MySQL query statement below
-SELECT
- e1.name AS Employee
+SELECT e1.name Employee
FROM
- Employee AS e1
- JOIN Employee AS e2 ON e1.managerId = e2.id
+ Employee e1
+ JOIN Employee e2 ON e1.managerId = e2.id
WHERE e1.salary > e2.salary;
```
diff --git a/solution/0100-0199/0181.Employees Earning More Than Their Managers/README_EN.md b/solution/0100-0199/0181.Employees Earning More Than Their Managers/README_EN.md
index 60ad47ffbcc60..4c5b427dbfe48 100644
--- a/solution/0100-0199/0181.Employees Earning More Than Their Managers/README_EN.md
+++ b/solution/0100-0199/0181.Employees Earning More Than Their Managers/README_EN.md
@@ -43,7 +43,7 @@ Each row of this table indicates the ID of an employee, their name, salary, and
Example 1:
-Input:
+Input:
Employee table:
+----+-------+--------+-----------+
| id | name | salary | managerId |
@@ -53,7 +53,7 @@ Employee table:
| 3 | Sam | 60000 | Null |
| 4 | Max | 90000 | Null |
+----+-------+--------+-----------+
-Output:
+Output:
+----------+
| Employee |
+----------+
@@ -68,7 +68,9 @@ Employee table:
-### Solution 1
+### Solution 1: Self-Join + Conditional Filtering
+
+We can find employees' salaries and their managers' salaries by self-joining the `Employee` table, then filter out employees whose salaries are higher than their managers' salaries.
@@ -79,44 +81,22 @@ import pandas as pd
def find_employees(employee: pd.DataFrame) -> pd.DataFrame:
- df = employee.merge(right=employee, how="left", left_on="managerId", right_on="id")
- emp = df[df["salary_x"] > df["salary_y"]]["name_x"]
-
- return pd.DataFrame({"Employee": emp})
-```
-
-#### MySQL
-
-```sql
-SELECT Name AS Employee
-FROM Employee AS Curr
-WHERE
- Salary > (
- SELECT Salary
- FROM Employee
- WHERE Id = Curr.ManagerId
- );
+ merged = employee.merge(
+ employee, left_on="managerId", right_on="id", suffixes=("", "_manager")
+ )
+ result = merged[merged["salary"] > merged["salary_manager"]][["name"]]
+ result.columns = ["Employee"]
+ return result
```
-
-
-
-
-
-
-### Solution 2
-
-
-
#### MySQL
```sql
# Write your MySQL query statement below
-SELECT
- e1.name AS Employee
+SELECT e1.name Employee
FROM
- Employee AS e1
- JOIN Employee AS e2 ON e1.managerId = e2.id
+ Employee e1
+ JOIN Employee e2 ON e1.managerId = e2.id
WHERE e1.salary > e2.salary;
```
diff --git a/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.py b/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.py
index 2084f018ba6ad..e9d9b508f46b9 100644
--- a/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.py
+++ b/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.py
@@ -2,7 +2,9 @@
def find_employees(employee: pd.DataFrame) -> pd.DataFrame:
- df = employee.merge(right=employee, how="left", left_on="managerId", right_on="id")
- emp = df[df["salary_x"] > df["salary_y"]]["name_x"]
-
- return pd.DataFrame({"Employee": emp})
+ merged = employee.merge(
+ employee, left_on="managerId", right_on="id", suffixes=("", "_manager")
+ )
+ result = merged[merged["salary"] > merged["salary_manager"]][["name"]]
+ result.columns = ["Employee"]
+ return result
diff --git a/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.sql b/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.sql
index da349ae2189b6..09530491aef0c 100644
--- a/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.sql
+++ b/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution.sql
@@ -1,8 +1,6 @@
-SELECT Name AS Employee
-FROM Employee AS Curr
-WHERE
- Salary > (
- SELECT Salary
- FROM Employee
- WHERE Id = Curr.ManagerId
- );
+# Write your MySQL query statement below
+SELECT e1.name Employee
+FROM
+ Employee e1
+ JOIN Employee e2 ON e1.managerId = e2.id
+WHERE e1.salary > e2.salary;
diff --git a/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution2.sql b/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution2.sql
deleted file mode 100644
index b285eab38d01a..0000000000000
--- a/solution/0100-0199/0181.Employees Earning More Than Their Managers/Solution2.sql
+++ /dev/null
@@ -1,7 +0,0 @@
-# Write your MySQL query statement below
-SELECT
- e1.name AS Employee
-FROM
- Employee AS e1
- JOIN Employee AS e2 ON e1.managerId = e2.id
-WHERE e1.salary > e2.salary;
diff --git a/solution/0400-0499/0459.Repeated Substring Pattern/README.md b/solution/0400-0499/0459.Repeated Substring Pattern/README.md
index cd3b2ad9c8bb8..3ed5096f5445e 100644
--- a/solution/0400-0499/0459.Repeated Substring Pattern/README.md
+++ b/solution/0400-0499/0459.Repeated Substring Pattern/README.md
@@ -131,39 +131,4 @@ impl Solution {
-
-
-### 方法二
-
-
-
-#### TypeScript
-
-```ts
-function repeatedSubstringPattern(s: string): boolean {
- const n = s.length;
- for (let i = 0; i < n >> 1; i++) {
- const len = i + 1;
- if (n % len !== 0) {
- continue;
- }
- const t = s.slice(0, len);
- let j: number;
- for (j = len; j < n; j += len) {
- if (s.slice(j, j + len) !== t) {
- break;
- }
- }
- if (j === n) {
- return true;
- }
- }
- return false;
-}
-```
-
-
-
-
-
diff --git a/solution/0400-0499/0459.Repeated Substring Pattern/README_EN.md b/solution/0400-0499/0459.Repeated Substring Pattern/README_EN.md
index 796b58a2b53c5..f537437ecaf1f 100644
--- a/solution/0400-0499/0459.Repeated Substring Pattern/README_EN.md
+++ b/solution/0400-0499/0459.Repeated Substring Pattern/README_EN.md
@@ -121,39 +121,4 @@ impl Solution {
-
-
-### Solution 2
-
-
-
-#### TypeScript
-
-```ts
-function repeatedSubstringPattern(s: string): boolean {
- const n = s.length;
- for (let i = 0; i < n >> 1; i++) {
- const len = i + 1;
- if (n % len !== 0) {
- continue;
- }
- const t = s.slice(0, len);
- let j: number;
- for (j = len; j < n; j += len) {
- if (s.slice(j, j + len) !== t) {
- break;
- }
- }
- if (j === n) {
- return true;
- }
- }
- return false;
-}
-```
-
-
-
-
-
diff --git a/solution/0400-0499/0459.Repeated Substring Pattern/Solution2.ts b/solution/0400-0499/0459.Repeated Substring Pattern/Solution2.ts
deleted file mode 100644
index 04bb2475a988e..0000000000000
--- a/solution/0400-0499/0459.Repeated Substring Pattern/Solution2.ts
+++ /dev/null
@@ -1,20 +0,0 @@
-function repeatedSubstringPattern(s: string): boolean {
- const n = s.length;
- for (let i = 0; i < n >> 1; i++) {
- const len = i + 1;
- if (n % len !== 0) {
- continue;
- }
- const t = s.slice(0, len);
- let j: number;
- for (j = len; j < n; j += len) {
- if (s.slice(j, j + len) !== t) {
- break;
- }
- }
- if (j === n) {
- return true;
- }
- }
- return false;
-}
diff --git a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README.md b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README.md
index 9766fcc59c830..ecaba2c407888 100644
--- a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README.md
+++ b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README.md
@@ -115,7 +115,9 @@ public:
sort(nums.begin(), nums.end());
int k = nums[nums.size() >> 1];
int ans = 0;
- for (int& v : nums) ans += abs(v - k);
+ for (int& v : nums) {
+ ans += abs(v - k);
+ }
return ans;
}
};
@@ -147,8 +149,8 @@ func abs(x int) int {
```ts
function minMoves2(nums: number[]): number {
nums.sort((a, b) => a - b);
- const mid = nums[nums.length >> 1];
- return nums.reduce((r, v) => r + Math.abs(v - mid), 0);
+ const k = nums[nums.length >> 1];
+ return nums.reduce((r, v) => r + Math.abs(v - k), 0);
}
```
@@ -158,12 +160,12 @@ function minMoves2(nums: number[]): number {
impl Solution {
pub fn min_moves2(mut nums: Vec) -> i32 {
nums.sort();
- let mid = nums[nums.len() / 2];
- let mut res = 0;
+ let k = nums[nums.len() / 2];
+ let mut ans = 0;
for num in nums.iter() {
- res += (num - mid).abs();
+ ans += (num - k).abs();
}
- res
+ ans
}
}
```
@@ -172,35 +174,4 @@ impl Solution {
-
-
-### 方法二：排序 + 前缀和
-
-如果我们不知道中位数的性质，也可以使用前缀和的方法来求解。
-
-时间复杂度 $O(n\log n)$，空间复杂度 $O(n)$。
-
-
-
-#### Python3
-
-```python
-class Solution:
- def minMoves2(self, nums: List[int]) -> int:
- def move(i):
- v = nums[i]
- a = v * i - s[i]
- b = s[-1] - s[i + 1] - v * (n - i - 1)
- return a + b
-
- nums.sort()
- s = [0] + list(accumulate(nums))
- n = len(nums)
- return min(move(i) for i in range(n))
-```
-
-
-
-
-
diff --git a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README_EN.md b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README_EN.md
index 30f763d29f493..86c2ff43fc2b4 100644
--- a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README_EN.md
+++ b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/README_EN.md
@@ -57,7 +57,24 @@ Only two moves are needed (remember each move increments or decrements one eleme
-### Solution 1
+### Solution 1: Sorting + Median
+
+This problem can be abstracted to finding a point on a number line such that the sum of distances from $n$ points to this point is minimized. The answer is the median of the $n$ points.
+
+The median has the property that the sum of distances from all numbers to the median is minimized.
+
+Proof:
+
+First, given a sorted sequence $a_1, a_2, \cdots, a_n$, we assume $x$ is the point that minimizes the sum of distances from $a_1$ to $a_n$. Clearly, $x$ must lie between $a_1$ and $a_n$. Since the distances from $a_1$ and $a_n$ to $x$ are equal and both equal to $a_n - a_1$, we can ignore $a_1$ and $a_n$ and only consider $a_2, a_3, \cdots, a_{n-1}$. This reduces the problem to finding a point within $a_2, a_3, \cdots, a_{n-1}$ that minimizes the sum of distances. By iterating this process, we conclude that the median of a sequence minimizes the sum of distances to the other numbers.
+
+In this problem, we can first sort the array, then find the median, and finally calculate the sum of distances from all numbers to the median.
+
+The time complexity is $O(n \log n)$, and the space complexity is $O(\log n)$.
+
+Similar problems:
+
+- [296. Best Meeting Point](https://github.com/doocs/leetcode/blob/main/solution/0200-0299/0296.Best%20Meeting%20Point/README_EN.md)
+- [2448. Minimum Cost to Make Array Equal](https://github.com/doocs/leetcode/blob/main/solution/2400-2499/2448.Minimum%20Cost%20to%20Make%20Array%20Equal/README_EN.md)
@@ -96,7 +113,9 @@ public:
sort(nums.begin(), nums.end());
int k = nums[nums.size() >> 1];
int ans = 0;
- for (int& v : nums) ans += abs(v - k);
+ for (int& v : nums) {
+ ans += abs(v - k);
+ }
return ans;
}
};
@@ -128,8 +147,8 @@ func abs(x int) int {
```ts
function minMoves2(nums: number[]): number {
nums.sort((a, b) => a - b);
- const mid = nums[nums.length >> 1];
- return nums.reduce((r, v) => r + Math.abs(v - mid), 0);
+ const k = nums[nums.length >> 1];
+ return nums.reduce((r, v) => r + Math.abs(v - k), 0);
}
```
@@ -139,12 +158,12 @@ function minMoves2(nums: number[]): number {
impl Solution {
pub fn min_moves2(mut nums: Vec) -> i32 {
nums.sort();
- let mid = nums[nums.len() / 2];
- let mut res = 0;
+ let k = nums[nums.len() / 2];
+ let mut ans = 0;
for num in nums.iter() {
- res += (num - mid).abs();
+ ans += (num - k).abs();
}
- res
+ ans
}
}
```
@@ -153,31 +172,4 @@ impl Solution {
-
-
-### Solution 2
-
-
-
-#### Python3
-
-```python
-class Solution:
- def minMoves2(self, nums: List[int]) -> int:
- def move(i):
- v = nums[i]
- a = v * i - s[i]
- b = s[-1] - s[i + 1] - v * (n - i - 1)
- return a + b
-
- nums.sort()
- s = [0] + list(accumulate(nums))
- n = len(nums)
- return min(move(i) for i in range(n))
-```
-
-
-
-
-
diff --git a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.cpp b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.cpp
index b51d3aa425ee1..3ea2196946003 100644
--- a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.cpp
+++ b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.cpp
@@ -4,7 +4,9 @@ class Solution {
sort(nums.begin(), nums.end());
int k = nums[nums.size() >> 1];
int ans = 0;
- for (int& v : nums) ans += abs(v - k);
+ for (int& v : nums) {
+ ans += abs(v - k);
+ }
return ans;
}
-};
\ No newline at end of file
+};
diff --git a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.rs b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.rs
index e561b2e2e178b..82b2a43d13e1c 100644
--- a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.rs
+++ b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.rs
@@ -1,11 +1,11 @@
impl Solution {
pub fn min_moves2(mut nums: Vec) -> i32 {
nums.sort();
- let mid = nums[nums.len() / 2];
- let mut res = 0;
+ let k = nums[nums.len() / 2];
+ let mut ans = 0;
for num in nums.iter() {
- res += (num - mid).abs();
+ ans += (num - k).abs();
}
- res
+ ans
}
}
diff --git a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.ts b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.ts
index f01325954ee85..b209cf697dd2d 100644
--- a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.ts
+++ b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution.ts
@@ -1,5 +1,5 @@
function minMoves2(nums: number[]): number {
nums.sort((a, b) => a - b);
- const mid = nums[nums.length >> 1];
- return nums.reduce((r, v) => r + Math.abs(v - mid), 0);
+ const k = nums[nums.length >> 1];
+ return nums.reduce((r, v) => r + Math.abs(v - k), 0);
}
diff --git a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution2.py b/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution2.py
deleted file mode 100644
index f099225ea8553..0000000000000
--- a/solution/0400-0499/0462.Minimum Moves to Equal Array Elements II/Solution2.py
+++ /dev/null
@@ -1,12 +0,0 @@
-class Solution:
- def minMoves2(self, nums: List[int]) -> int:
- def move(i):
- v = nums[i]
- a = v * i - s[i]
- b = s[-1] - s[i + 1] - v * (n - i - 1)
- return a + b
-
- nums.sort()
- s = [0] + list(accumulate(nums))
- n = len(nums)
- return min(move(i) for i in range(n))