[pull] main from doocs:main

lcp/LCP 61. 气温变化趋势/README.md

@@ -40,22 +40,105 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：动态规划**
+
+我们用变量 $f$ 维护当前趋势相同的连续天数，用变量 $ans$ 维护最大的连续天数。
+
+遍历数组，对于第 $i$ 天，记两地的气温变化趋势分别为 $x$ 和 $y$，如果 $x$ 和 $y$ 均为 $0$ 或者 $x$ 和 $y$ 均为正数或负数，则说明第 $i$ 天和第 $i+1$ 天的气温变化趋势相同，此时 $f$ 自增 $1$，并更新 $ans$；否则说明第 $i$ 天和第 $i+1$ 天的气温变化趋势不同，此时 $f$ 重置为 $0$。
+
+最终返回 $ans$ 即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def temperatureTrend(self, temperatureA: List[int], temperatureB: List[int]) -> int:
+        ans = f = 0
+        n = len(temperatureA)
+        for i in range(n - 1):
+            x = temperatureA[i + 1] - temperatureA[i]
+            y = temperatureB[i + 1] - temperatureB[i]
+            if x == y == 0 or x * y > 0:
+                f += 1
+                ans = max(ans, f)
+            else:
+                f = 0
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int temperatureTrend(int[] temperatureA, int[] temperatureB) {
+        int ans = 0, f = 0;
+        for (int i = 0; i < temperatureA.length - 1; ++i) {
+            int x = temperatureA[i + 1] - temperatureA[i];
+            int y = temperatureB[i + 1] - temperatureB[i];
+            if ((x == 0 && y == 0) || x * y > 0) {
+                ans = Math.max(ans, ++f);
+            } else {
+                f = 0;
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int temperatureTrend(vector<int>& temperatureA, vector<int>& temperatureB) {
+        int ans = 0, f = 0;
+        for (int i = 0; i < temperatureA.size() - 1; ++i) {
+            int x = temperatureA[i + 1] - temperatureA[i];
+            int y = temperatureB[i + 1] - temperatureB[i];
+            if ((x == 0 && y == 0) || x * y > 0) {
+                ans = max(ans, ++f);
+            } else {
+                f = 0;
+            }
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func temperatureTrend(temperatureA []int, temperatureB []int) int {
+	ans, f := 0, 0
+	for i := range temperatureA[1:] {
+		x := temperatureA[i+1] - temperatureA[i]
+		y := temperatureB[i+1] - temperatureB[i]
+		if (x == 0 && y == 0) || x*y > 0 {
+			f++
+			ans = max(ans, f)
+		} else {
+			f = 0
+		}
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **...**

lcp/LCP 61. 气温变化趋势/Solution.cpp

@@ -0,0 +1,16 @@
+class Solution {
+public:
+    int temperatureTrend(vector<int>& temperatureA, vector<int>& temperatureB) {
+        int ans = 0, f = 0;
+        for (int i = 0; i < temperatureA.size() - 1; ++i) {
+            int x = temperatureA[i + 1] - temperatureA[i];
+            int y = temperatureB[i + 1] - temperatureB[i];
+            if ((x == 0 && y == 0) || x * y > 0) {
+                ans = max(ans, ++f);
+            } else {
+                f = 0;
+            }
+        }
+        return ans;
+    }
+};

lcp/LCP 61. 气温变化趋势/Solution.go

@@ -0,0 +1,21 @@
+func temperatureTrend(temperatureA []int, temperatureB []int) int {
+	ans, f := 0, 0
+	for i := range temperatureA[1:] {
+		x := temperatureA[i+1] - temperatureA[i]
+		y := temperatureB[i+1] - temperatureB[i]
+		if (x == 0 && y == 0) || x*y > 0 {
+			f++
+			ans = max(ans, f)
+		} else {
+			f = 0
+		}
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}

lcp/LCP 61. 气温变化趋势/Solution.java

@@ -0,0 +1,15 @@
+class Solution {
+    public int temperatureTrend(int[] temperatureA, int[] temperatureB) {
+        int ans = 0, f = 0;
+        for (int i = 0; i < temperatureA.length - 1; ++i) {
+            int x = temperatureA[i + 1] - temperatureA[i];
+            int y = temperatureB[i + 1] - temperatureB[i];
+            if ((x == 0 && y == 0) || x * y > 0) {
+                ans = Math.max(ans, ++f);
+            } else {
+                f = 0;
+            }
+        }
+        return ans;
+    }
+}

lcp/LCP 61. 气温变化趋势/Solution.py

@@ -0,0 +1,13 @@
+class Solution:
+    def temperatureTrend(self, temperatureA: List[int], temperatureB: List[int]) -> int:
+        ans = f = 0
+        n = len(temperatureA)
+        for i in range(n - 1):
+            x = temperatureA[i + 1] - temperatureA[i]
+            y = temperatureB[i + 1] - temperatureB[i]
+            if x == y == 0 or x * y > 0:
+                f += 1
+                ans = max(ans, f)
+            else:
+                f = 0
+        return ans

lcp/LCP 62. 交通枢纽/README.md

@@ -46,22 +46,135 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：统计入度和出度**
+
+我们创建两个数组 $ind$ 和 $outd$，分别用于记录每个点的入度和出度，用哈希表 $s$ 保存每个节点。
+
+接下来，遍历每个节点 $c$，如果存在 $ind[c]$ 等于节点总数减去 $1$，并且 $outd[c]=0$，说明存在满足条件的交通枢纽，返回 $c$。
+
+否则遍历结束，返回 $-1$。
+
+时间复杂度 $O(n + m)$，空间复杂度 $O(n)$。其中 $n$ 和 $m$ 分别是节点数量以及路径的数量。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def transportationHub(self, path: List[List[int]]) -> int:
+        ind = Counter()
+        outd = Counter()
+        s = set()
+        for a, b in path:
+            s.add(a)
+            s.add(b)
+            outd[a] += 1
+            ind[b] += 1
+        for c in s:
+            if ind[c] == len(s) - 1 and outd[c] == 0:
+                return c
+        return -1
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int transportationHub(int[][] path) {
+        int[] ind = new int[1001];
+        int[] outd = new int[1001];
+        Set<Integer> s = new HashSet<>();
+        for (int[] p : path) {
+            int a = p[0], b = p[1];
+            s.add(a);
+            s.add(b);
+            ind[b]++;
+            outd[a]++;
+        }
+        for (int c : s) {
+            if (ind[c] == s.size() - 1 && outd[c] == 0) {
+                return c;
+            }
+        }
+        return -1;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int transportationHub(vector<vector<int>>& path) {
+        int ind[1001]{};
+        int outd[1001]{};
+        unordered_set<int> s;
+        for (auto& p : path) {
+            int a = p[0], b = p[1];
+            s.insert(a);
+            s.insert(b);
+            ind[b]++;
+            outd[a]++;
+        }
+        for (int c : s) {
+            if (ind[c] == s.size() - 1 && outd[c] == 0) {
+                return c;
+            }
+        }
+        return -1;
+    }
+};
+```
+
+### **Go**
+
+```go
+func transportationHub(path [][]int) int {
+	ind := [1001]int{}
+	outd := [1001]int{}
+	s := map[int]struct{}{}
+	for _, p := range path {
+		a, b := p[0], p[1]
+		s[a] = struct{}{}
+		s[b] = struct{}{}
+		outd[a]++
+		ind[b]++
+	}
+	for c := range s {
+		if ind[c] == len(s)-1 && outd[c] == 0 {
+			return c
+		}
+	}
+	return -1
+}
+```
 
+### **TypeScript**
+
+```ts
+function transportationHub(path: number[][]): number {
+    const ind: number[] = new Array(1001).fill(0);
+    const outd: number[] = new Array(1001).fill(0);
+    const s: Set<number> = new Set();
+    for (const [a, b] of path) {
+        s.add(a);
+        s.add(b);
+        ind[b]++;
+        outd[a]++;
+    }
+    for (const c of s) {
+        if (ind[c] === s.size - 1 && outd[c] === 0) {
+            return c;
+        }
+    }
+    return -1;
+}
 ```
 
 ### **...**

lcp/LCP 62. 交通枢纽/Solution.cpp

@@ -0,0 +1,21 @@
+class Solution {
+public:
+    int transportationHub(vector<vector<int>>& path) {
+        int ind[1001]{};
+        int outd[1001]{};
+        unordered_set<int> s;
+        for (auto& p : path) {
+            int a = p[0], b = p[1];
+            s.insert(a);
+            s.insert(b);
+            ind[b]++;
+            outd[a]++;
+        }
+        for (int c : s) {
+            if (ind[c] == s.size() - 1 && outd[c] == 0) {
+                return c;
+            }
+        }
+        return -1;
+    }
+};

lcp/LCP 62. 交通枢纽/Solution.go

@@ -0,0 +1,18 @@
+func transportationHub(path [][]int) int {
+	ind := [1001]int{}
+	outd := [1001]int{}
+	s := map[int]struct{}{}
+	for _, p := range path {
+		a, b := p[0], p[1]
+		s[a] = struct{}{}
+		s[b] = struct{}{}
+		outd[a]++
+		ind[b]++
+	}
+	for c := range s {
+		if ind[c] == len(s)-1 && outd[c] == 0 {
+			return c
+		}
+	}
+	return -1
+}

lcp/LCP 62. 交通枢纽/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public int transportationHub(int[][] path) {
+        int[] ind = new int[1001];
+        int[] outd = new int[1001];
+        Set<Integer> s = new HashSet<>();
+        for (int[] p : path) {
+            int a = p[0], b = p[1];
+            s.add(a);
+            s.add(b);
+            ind[b]++;
+            outd[a]++;
+        }
+        for (int c : s) {
+            if (ind[c] == s.size() - 1 && outd[c] == 0) {
+                return c;
+            }
+        }
+        return -1;
+    }
+}

lcp/LCP 62. 交通枢纽/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def transportationHub(self, path: List[List[int]]) -> int:
+        ind = Counter()
+        outd = Counter()
+        s = set()
+        for a, b in path:
+            s.add(a)
+            s.add(b)
+            outd[a] += 1
+            ind[b] += 1
+        for c in s:
+            if ind[c] == len(s) - 1 and outd[c] == 0:
+                return c
+        return -1