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feat: add solutions to lcof2 problem: No.074
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‎lcof2/剑指 Offer II 074. 合并区间/README.md

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<!-- 这里可写通用的实现逻辑 -->
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区间合并，将所有存在交集的区间进行合并。
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模板：
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```py
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def merge(intervals):
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res = []
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intervals.sort(key=lambda x: x[0])
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st = ed = -1
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for s, e in intervals:
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if ed < s:
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if st != -1:
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res.append([st, ed])
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st, ed = e[0], e[1]
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else:
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ed = max(ed, e[1])
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if st != -1:
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res.append([st, ed])
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return res
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```
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def merge(self, intervals: List[List[int]]) -> List[List[int]]:
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intervals.sort(key=lambda x: x[0])
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st = ed = -1
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res = []
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for s, e in intervals:
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if ed < s:
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if st != -1:
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res.append([st, ed])
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st, ed = s, e
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else:
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ed = max(ed, e)
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if st != -1:
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res.append([st, ed])
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return res
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int[][] merge(int[][] intervals) {
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Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
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int st = -1, ed = -1;
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List<int[]> res = new ArrayList<>();
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for (int[] e : intervals) {
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if (ed < e[0]) {
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if (st != -1) {
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res.add(new int[]{st, ed});
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}
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st = e[0];
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ed = e[1];
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} else {
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ed = Math.max(ed, e[1]);
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}
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}
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if (st != -1) {
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res.add(new int[]{st, ed});
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}
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return res.toArray(new int[res.size()][]);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<vector<int>> merge(vector<vector<int>> &intervals) {
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sort(intervals.begin(), intervals.end());
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vector<vector<int>> res;
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int st = -1, ed = -1;
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for (auto e : intervals)
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{
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if (ed < e[0])
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{
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if (st != -1)
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{
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res.push_back({st, ed});
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}
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st = e[0];
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ed = e[1];
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}
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else
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{
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ed = max(ed, e[1]);
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}
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}
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if (st != -1)
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{
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res.push_back({st, ed});
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}
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return res;
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}
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};
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```
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### **Go**
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```go
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func merge(intervals [][]int) [][]int {
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var res [][]int
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sort.Slice(intervals, func(i, j int) bool {
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return intervals[i][0] < intervals[j][0]
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})
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st, ed := -1, -1
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for _, e := range intervals {
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if ed < e[0] {
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if st != -1 {
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res = append(res, []int{st, ed})
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}
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st, ed = e[0], e[1]
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} else {
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ed = max(ed, e[1])
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}
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}
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if st != -1 {
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res = append(res, []int{st, ed})
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}
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return res
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **C#**
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```cpp
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public class Solution {
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public int[][] Merge(int[][] intervals) {
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var res = new List<int[]>();
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int st = -1, ed = -1;
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foreach (var e in intervals.OrderBy(a => a[0]))
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{
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if (ed < e[0])
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{
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if (st != -1)
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{
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res.Add(new int[] { st, ed });
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}
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st = e[0];
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ed = e[1];
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}
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else
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{
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ed = Math.Max(ed, e[1]);
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}
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}
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if (st != -1)
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{
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res.Add(new int[] { st, ed });
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}
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return res.ToArray();
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}
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}
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```
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### **...**

‎lcof2/剑指 Offer II 074. 合并区间/Solution.cpp

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class Solution {
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public:
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vector<vector<int>> merge(vector<vector<int>> &intervals) {
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sort(intervals.begin(), intervals.end());
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vector<vector<int>> res;
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int st = -1, ed = -1;
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for (auto e : intervals)
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{
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if (ed < e[0])
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{
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if (st != -1)
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{
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res.push_back({st, ed});
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}
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st = e[0];
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ed = e[1];
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}
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else
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{
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ed = max(ed, e[1]);
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}
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}
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if (st != -1)
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{
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res.push_back({st, ed});
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}
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return res;
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}
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};

‎lcof2/剑指 Offer II 074. 合并区间/Solution.cs

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public class Solution {
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public int[][] Merge(int[][] intervals) {
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var res = new List<int[]>();
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int st = -1, ed = -1;
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foreach (var e in intervals.OrderBy(a => a[0]))
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{
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if (ed < e[0])
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{
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if (st != -1)
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{
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res.Add(new int[] { st, ed });
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}
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st = e[0];
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ed = e[1];
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}
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else
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{
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ed = Math.Max(ed, e[1]);
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}
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}
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if (st != -1)
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{
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res.Add(new int[] { st, ed });
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}
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return res.ToArray();
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}
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}

‎lcof2/剑指 Offer II 074. 合并区间/Solution.go

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func merge(intervals [][]int) [][]int {
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var res [][]int
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sort.Slice(intervals, func(i, j int) bool {
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return intervals[i][0] < intervals[j][0]
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})
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st, ed := -1, -1
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for _, e := range intervals {
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if ed < e[0] {
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if st != -1 {
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res = append(res, []int{st, ed})
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}
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st, ed = e[0], e[1]
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} else {
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ed = max(ed, e[1])
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}
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}
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if st != -1 {
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res = append(res, []int{st, ed})
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}
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return res
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎lcof2/剑指 Offer II 074. 合并区间/Solution.java

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class Solution {
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public int[][] merge(int[][] intervals) {
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Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
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int st = -1, ed = -1;
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List<int[]> res = new ArrayList<>();
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for (int[] e : intervals) {
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if (ed < e[0]) {
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if (st != -1) {
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res.add(new int[]{st, ed});
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}
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st = e[0];
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ed = e[1];
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} else {
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ed = Math.max(ed, e[1]);
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}
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}
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if (st != -1) {
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res.add(new int[]{st, ed});
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}
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return res.toArray(new int[res.size()][]);
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}
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}

‎lcof2/剑指 Offer II 074. 合并区间/Solution.py

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class Solution:
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def merge(self, intervals: List[List[int]]) -> List[List[int]]:
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intervals.sort(key=lambda x: x[0])
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st = ed = -1
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res = []
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for s, e in intervals:
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if ed < s:
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if st != -1:
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res.append([st, ed])
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st, ed = s, e
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else:
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ed = max(ed, e)
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if st != -1:
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res.append([st, ed])
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return res

‎solution/1900-1999/1978.Employees Whose Manager Left the Company/README.md

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# [1978. Employees Whose Manager Left the Company](https://leetcode-cn.com/problems/employees-whose-manager-left-the-company)
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[English Version](/solution/1900-1999/1978.Employees%20Whose%20Manager%20Left%20the%20Company/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>Table: <code>Employees</code></p>
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<pre>
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+-------------+----------+
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| Column Name | Type |
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+-------------+----------+
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| employee_id | int |
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| name | varchar |
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| manager_id | int |
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| salary | int |
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+-------------+----------+
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employee_id is the primary key for this table.
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This table contains information about the employees, their salary, and the ID of their manager. Some employees do not have a manager (manager_id is null).
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</pre>
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<p>&nbsp;</p>
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<p>Write an SQL query to report the IDs of the employees whose salary is strictly less than <code>$30000</code> and whose manager left the company. When a manager leaves the company, their information is deleted from the <code>Employees</code> table, but the reports still have their <code>manager_id</code> set to the manager that left.</p>
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<p>Return the result table ordered by <code>employee_id</code>.</p>
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<p>The query result format is in the following example.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input: </strong>
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Employees table:
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+-------------+-----------+------------+--------+
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| employee_id | name | manager_id | salary |
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+-------------+-----------+------------+--------+
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| 3 | Mila | 9 | 60301 |
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| 12 | Antonella | null | 31000 |
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| 13 | Emery | null | 67084 |
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| 1 | Kalel | 11 | 21241 |
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| 9 | Mikaela | null | 50937 |
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| 11 | Joziah | 6 | 28485 |
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+-------------+-----------+------------+--------+
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<strong>Output:</strong>
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+-------------+
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| employee_id |
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+-------------+
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| 11 |
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+-------------+
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<strong>Explanation:</strong>
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The employees with a salary less than $30000 are 1 (Kalel) and 11 (Joziah).
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Kalel&#39;s manager is employee 11, who is still in the company (Joziah).
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Joziah&#39;s manager is employee 6, who left the company because there is no row for employee 6 as it was deleted.
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</pre>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **SQL**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```sql
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```
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<!-- tabs:end -->

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